Okay, so I need to find the distance between two parallel lines in 3D space. Hmm, let me recall how to do this. I think the formula involves the cross product of the direction vectors and the vector connecting a point from each line. Let me try to remember the exact steps.

First, since the lines are parallel, their direction vectors must be scalar multiples of each other. Let me check that. The first line has direction vector (2, -9, -2) and the second one has (4, -18, -4). If I multiply the first direction vector by 2, I get (4, -18, -4), which is exactly the second direction vector. So yes, they are parallel. Good.

Now, to find the distance between two parallel lines, the formula is the magnitude of the cross product of the direction vectors divided by the magnitude of the direction vector, multiplied by the magnitude of the vector connecting a point on the first line to a point on the second line. Wait, maybe I should write it down more clearly.

The formula is: distance = |(P2 - P1) · (v × w)| / |v × w|, where P1 and P2 are points on each line, and v and w are their direction vectors. Wait, no, that's not quite right. Let me think again. The distance between two parallel lines can be found by taking the length of the projection of the vector connecting any two points on the lines onto a vector perpendicular to both lines, which would be the cross product of their direction vectors.

So the distance is |(P2 - P1) · (v × w)| / |v × w|. Wait, but I think it's actually |(P2 - P1) × direction| divided by |direction|, but since they are parallel, direction vectors are the same. Hmm, maybe I should derive it.

Alternatively, another approach is to take a vector connecting a point on the first line to a point on the second line, say vector AP, where A is a point on the first line and P is a point on the second line. Then the distance is the length of the component of AP that's perpendicular to the direction vector of the lines. Since the lines are parallel, the distance can be found by |AP × direction| divided by |direction|.

Yes, that makes sense. So if I take the cross product of the vector AP and the direction vector, and then divide by the magnitude of the direction vector, that should give me the distance.

Let me write down the steps:

1. Identify points on each line. Let's take t=0 for the first line, so point A is (3, -10, 1). For the second line, take u=0, so point B is (-5, -3, 6).

2. Compute the vector AP = B - A = (-5 - 3, -3 - (-10), 6 - 1) = (-8, 7, 5).

3. The direction vector of the lines is v = (2, -9, -2).

4. Compute the cross product of AP and v. Let's do that:

AP × v = |i   j   k|
         |-8   7   5|
         |2   -9  -2|

Calculating the determinant:

i*(7*(-2) - 5*(-9)) - j*(-8*(-2) - 5*2) + k*(-8*(-9) - 7*2)

Compute each component:

i*( -14 + 45 ) = i*(31)

-j*(16 - 10) = -j*(6)

k*(72 - 14) = k*(58)

So AP × v = (31, -6, 58)

5. Find the magnitude of this cross product:

|AP × v| = sqrt(31² + (-6)² + 58²) = sqrt(961 + 36 + 3364) = sqrt(961 + 36 is 997, 997 + 3364 is 4361). So sqrt(4361). Let me check that 4361 is correct. 31^2 is 961, 6^2 is 36, 58^2 is 3364. Yes, 961 + 36 = 997, 997 + 3364 = 4361. So sqrt(4361). Hmm, is 4361 a perfect square? Let me check. 66^2 is 4356, 67^2 is 4489. So no, it's between 66 and 67. So sqrt(4361) is approximately 66.03, but we can leave it as sqrt(4361) for now.

6. The magnitude of the direction vector v is |v| = sqrt(2² + (-9)^2 + (-2)^2) = sqrt(4 + 81 + 4) = sqrt(89). So sqrt(89) is approximately 9.433.

7. Then the distance is |AP × v| / |v| = sqrt(4361)/sqrt(89). Let's compute that. sqrt(4361)/sqrt(89) = sqrt(4361/89). Let's divide 4361 by 89. 89*49 = 4361 because 89*50=4450, minus 89 is 4361. So 4361/89 = 49. Therefore, sqrt(49) = 7. Wait, that's nice! So the distance is 7. That's the answer. Wait, really?

Let me check again. So cross product of AP and v is (31, -6, 58). Then |AP × v| is sqrt(31² + (-6)^2 + 58²) = sqrt(961 + 36 + 3364) = sqrt(4361). Then |v| is sqrt(89). Then 4361 divided by 89 is indeed 49, so sqrt(49) is 7. So the distance is 7. That's a whole number. That seems clean. So maybe I did this right.

But let me double-check each step to make sure I didn't make a mistake.

First, points A and B: A is (3, -10, 1) at t=0, B is (-5, -3, 6) at u=0. So vector AP is B - A = (-5-3, -3 - (-10), 6 -1) = (-8, 7, 5). Correct.

Cross product calculation:

i*(7*(-2) -5*(-9)) = i*(-14 +45) = i*31

-j*(-8*(-2) -5*2) = -j*(16 -10) = -j*6

k*(-8*(-9) -7*2) = k*(72 -14) = k*58. Correct.

So cross product is (31, -6, 58). Correct.

Magnitude squared: 31² = 961, (-6)^2=36, 58²=3364. Sum is 961+36=997, 997+3364=4361. Correct.

Divide by |v|, which is sqrt(89). Then 4361 divided by 89. Let's compute 89*49. 89*50=4450, so 89*49=4450-89=4361. Yes, exactly. So 4361/89=49. Therefore sqrt(49)=7. Correct.

Therefore, the distance is 7. That seems right. So I think that's the answer.

But just to be thorough, let me think if there's another way to approach this problem. For example, using parametric equations and minimizing the distance between a general point on each line.

Let me try that method to confirm.

First, parametrize both lines:

Line 1: (3, -10, 1) + t(2, -9, -2)

Line 2: (-5, -3, 6) + u(4, -18, -4)

Since the lines are parallel, the distance between them is the minimal distance between any two points on the lines. Let's denote a general point on Line 1 as P(t) = (3 + 2t, -10 -9t, 1 -2t)

And a general point on Line 2 as Q(u) = (-5 +4u, -3 -18u, 6 -4u)

The vector PQ = Q(u) - P(t) = (-5 +4u -3 -2t, -3 -18u - (-10 -9t), 6 -4u -1 +2t)

Simplify:

x-component: -8 +4u -2t

y-component: 7 -18u +9t

z-component: 5 -4u +2t

The distance squared between P(t) and Q(u) is the sum of the squares of these components:

D² = (-8 +4u -2t)^2 + (7 -18u +9t)^2 + (5 -4u +2t)^2

To find the minimal distance, we need to minimize D² with respect to t and u.

To do this, take partial derivatives with respect to t and u, set them equal to zero, and solve the system of equations.

First, compute the partial derivative with respect to t:

d(D²)/dt = 2*(-8 +4u -2t)*(-2) + 2*(7 -18u +9t)*(9) + 2*(5 -4u +2t)*(2)

Set this equal to zero:

-2*(-8 +4u -2t) + 9*(7 -18u +9t) + 2*(5 -4u +2t) = 0

Similarly, partial derivative with respect to u:

d(D²)/du = 2*(-8 +4u -2t)*(4) + 2*(7 -18u +9t)*(-18) + 2*(5 -4u +2t)*(-4)

Set this equal to zero:

4*(-8 +4u -2t) -18*(7 -18u +9t) -4*(5 -4u +2t) = 0

Now, let's compute these equations step by step.

First, the equation from partial derivative with respect to t:

-2*(-8 +4u -2t) + 9*(7 -18u +9t) + 2*(5 -4u +2t) = 0

Compute each term:

First term: -2*(-8 +4u -2t) = 16 -8u +4t

Second term: 9*(7 -18u +9t) = 63 -162u +81t

Third term: 2*(5 -4u +2t) = 10 -8u +4t

Add them together:

16 -8u +4t +63 -162u +81t +10 -8u +4t

Combine like terms:

Constants: 16 +63 +10 = 89

u terms: -8u -162u -8u = -178u

t terms: 4t +81t +4t = 89t

So equation becomes:

89 -178u +89t = 0

Divide both sides by 89:

1 - 2u + t = 0 => t = 2u -1

Now, the equation from partial derivative with respect to u:

4*(-8 +4u -2t) -18*(7 -18u +9t) -4*(5 -4u +2t) = 0

Compute each term:

First term: 4*(-8 +4u -2t) = -32 +16u -8t

Second term: -18*(7 -18u +9t) = -126 +324u -162t

Third term: -4*(5 -4u +2t) = -20 +16u -8t

Add them together:

-32 +16u -8t -126 +324u -162t -20 +16u -8t

Combine like terms:

Constants: -32 -126 -20 = -178

u terms: 16u +324u +16u = 356u

t terms: -8t -162t -8t = -178t

So equation becomes:

-178 +356u -178t = 0

Divide both sides by -178:

1 -2u + t = 0 => t = 2u -1

Wait, that's the same equation as before. So both partial derivatives lead to t = 2u -1. That means this is the only equation we have. But we need another equation to solve for u and t. Wait, but we have only one equation here. That suggests that perhaps the system is dependent, but actually, since we have only one independent equation, but we have two variables. Wait, but in reality, the minimal distance occurs when the vector PQ is perpendicular to both direction vectors of the lines. Since the lines are parallel, the direction vector is the same, so PQ should be perpendicular to the direction vector v.

Therefore, another condition is that PQ · v = 0. Let's check that.

Compute PQ · v = 0.

PQ is ( -8 +4u -2t, 7 -18u +9t, 5 -4u +2t )

v is (2, -9, -2)

Dot product:

2*(-8 +4u -2t) + (-9)*(7 -18u +9t) + (-2)*(5 -4u +2t) = 0

Compute each term:

First term: 2*(-8 +4u -2t) = -16 +8u -4t

Second term: -9*(7 -18u +9t) = -63 +162u -81t

Third term: -2*(5 -4u +2t) = -10 +8u -4t

Add them together:

-16 +8u -4t -63 +162u -81t -10 +8u -4t

Combine like terms:

Constants: -16 -63 -10 = -89

u terms: 8u +162u +8u = 178u

t terms: -4t -81t -4t = -89t

So equation becomes:

-89 +178u -89t = 0

Divide both sides by -89:

1 -2u + t = 0 => t = 2u -1

Same equation again. So all three conditions lead to the same equation, which is t = 2u -1. Therefore, we have a one-dimensional system. But how do we find the specific values of u and t?

Wait, perhaps we need to use the parametric equations to express the distance squared and then substitute t in terms of u, and then find the derivative with respect to u.

Alternatively, since the minimal distance occurs when PQ is perpendicular to the direction vector, which gives us the equation t = 2u -1. Then, we can substitute this into the expressions for P(t) and Q(u) and compute the distance.

Let me try that.

Given t = 2u -1, substitute into Q(u):

Q(u) = (-5 +4u, -3 -18u, 6 -4u)

And P(t) = (3 +2t, -10 -9t, 1 -2t) = (3 +2*(2u -1), -10 -9*(2u -1), 1 -2*(2u -1))

Compute each component:

x: 3 +4u -2 = 1 +4u

y: -10 -18u +9 = -1 -18u

z: 1 -4u +2 = 3 -4u

So P(t) is (1 +4u, -1 -18u, 3 -4u)

Now, vector PQ = Q(u) - P(t) = (-5 +4u - (1 +4u), -3 -18u - (-1 -18u), 6 -4u - (3 -4u))

Simplify each component:

x: -5 +4u -1 -4u = -6

y: -3 -18u +1 +18u = -2

z: 6 -4u -3 +4u = 3

Wait, so PQ is (-6, -2, 3). So the vector PQ is constant? That can't be right. Wait, but if t is expressed in terms of u, then PQ should depend on u. But here, after substitution, it's constant? That suggests that the minimal distance is achieved for any u? That doesn't make sense. Maybe I made a mistake here.

Wait, let's check the calculations again.

Compute P(t):

x: 3 +2t = 3 +2*(2u -1) = 3 +4u -2 = 1 +4u

y: -10 -9t = -10 -9*(2u -1) = -10 -18u +9 = -1 -18u

z: 1 -2t = 1 -2*(2u -1) = 1 -4u +2 = 3 -4u

Okay, that's correct.

Q(u):

x: -5 +4u

y: -3 -18u

z: 6 -4u

So PQ = Q(u) - P(t):

x: (-5 +4u) - (1 +4u) = -5 -1 +4u -4u = -6

y: (-3 -18u) - (-1 -18u) = -3 +1 -18u +18u = -2

z: (6 -4u) - (3 -4u) = 6 -3 -4u +4u = 3

So PQ is indeed (-6, -2, 3). So the vector PQ is constant, regardless of u? That implies that for any value of u, if t is set to 2u -1, the vector connecting the two points is the same. Which suggests that the lines are offset by this vector, and hence the distance is the magnitude of this vector. But wait, that would mean the distance is fixed, but earlier we found it to be 7, which is the magnitude of this vector.

Wait, let's compute the magnitude of (-6, -2, 3):

sqrt((-6)^2 + (-2)^2 +3^2) = sqrt(36 +4 +9) = sqrt(49) =7. So the distance is indeed 7. So that's the same answer as before.

Therefore, even though we approached it through minimizing the distance squared function, we end up with the same result. Therefore, the distance between the two parallel lines is 7. That seems to confirm the answer.

Alternatively, another method to compute the distance is using the formula:

Distance = |(b - a) · (v × w)| / |v × w|

But since the lines are parallel, their direction vectors v and w are scalar multiples, so v × w = 0. Wait, that's not possible. Wait, no, cross product of two parallel vectors is zero. So that formula isn't applicable here. Wait, but in our first approach, we used the cross product of the vector connecting the points and the direction vector. So maybe that formula is different.

Wait, no, the correct formula when the lines are not necessarily parallel is |(b - a) · (v × w)| / |v × w|. But when lines are parallel, v × w = 0, so that formula isn't helpful. Hence, instead, we compute the cross product of (b - a) with the direction vector, then divide by the magnitude of the direction vector. Which is exactly what we did.

So in this case, since the lines are parallel, the cross product is non-zero, and we can compute the distance as |(b - a) × v| / |v|. That gives us 7. So that's consistent.

Therefore, after verifying through multiple methods—using the cross product formula directly, and also by parametrizing the lines and minimizing the distance squared—we arrive at the same answer. Therefore, I can be confident that the distance between the two lines is 7.

**Final Answer**
\boxed{7}
Therefore, after carefully analyzing the problem, computing the cross product of the vector connecting a point on each line with the direction vector of the lines, and verifying through multiple methods, we find that the distance between the two parallel lines is $\boxed{7}$.
</think>