Okay, so I need to find a vector v such that when I take the cross product of (1, 2, -5) with v, I get (90, 30, 30). And among all possible such vectors, I have to find the one with the smallest magnitude. Hmm, cross products... right. Let me recall how cross products work.

First, cross product of two vectors a and b is another vector that's perpendicular to both a and b. Its magnitude is |a||b|sin(theta), where theta is the angle between them, and its direction is determined by the right-hand rule. But here, I need to solve for v given that the cross product equals a specific vector. So this is like solving a vector equation.

Let me denote the given vectors. Let a = (1, 2, -5), and the result is c = (90, 30, 30). So we have a × v = c. We need to find v. Wait, but cross product is not commutative, so a × v = c is different from v × a = c. So order matters here.

So, if a × v = c, then we can write this equation as a system of equations. Let me write the cross product formula. If a = (a1, a2, a3) and v = (v1, v2, v3), then a × v is:

(a2v3 - a3v2, a3v1 - a1v3, a1v2 - a2v1)

So setting this equal to c = (90, 30, 30), we get three equations:

1. a2v3 - a3v2 = 90
2. a3v1 - a1v3 = 30
3. a1v2 - a2v1 = 30

Plugging in the values from a = (1, 2, -5):

1. 2v3 - (-5)v2 = 90 → 2v3 + 5v2 = 90
2. (-5)v1 - 1v3 = 30 → -5v1 - v3 = 30
3. 1v2 - 2v1 = 30 → v2 - 2v1 = 30

So now we have a system of three equations with three variables: v1, v2, v3.

Let me write them again:

Equation 1: 2v3 + 5v2 = 90

Equation 2: -5v1 - v3 = 30

Equation 3: v2 - 2v1 = 30

So we have three equations. Let's see if we can solve this system. Let me try to express variables in terms of others.

From Equation 3: v2 = 30 + 2v1. Let's substitute this into Equations 1 and 2.

Substituting v2 into Equation 1:

2v3 + 5*(30 + 2v1) = 90

Compute that: 2v3 + 150 + 10v1 = 90 → 2v3 + 10v1 = 90 - 150 → 2v3 + 10v1 = -60

Divide both sides by 2: v3 + 5v1 = -30 → v3 = -30 -5v1

Now substitute v3 into Equation 2:

-5v1 - (-30 -5v1) = 30

Simplify: -5v1 +30 +5v1 = 30 → 30 = 30

Wait, that's 30=30, which is always true. So that means that Equations 2 and 3 are dependent, and we have infinitely many solutions. Which makes sense because the problem statement says there are infinitely many vectors v satisfying the cross product equation. The solutions form a line in 3D space.

So the general solution depends on one parameter. Let's see. From Equation 3: v2 = 30 + 2v1

From the above substitution, we have v3 = -30 -5v1

So if we let v1 = t, where t is a parameter, then:

v2 = 30 + 2t

v3 = -30 -5t

So the vector v can be written as:

v = (t, 30 + 2t, -30 -5t)

Now, we need to find the vector v with the smallest magnitude. The magnitude squared of v is:

||v||² = t² + (30 + 2t)² + (-30 -5t)²

We can expand this and find the value of t that minimizes this quadratic expression. Then take the square root to get the minimal magnitude.

Let me compute each term:

First term: t²

Second term: (30 + 2t)² = 900 + 120t + 4t²

Third term: (-30 -5t)² = (30 +5t)² = 900 + 300t +25t²

Adding them all together:

t² + 900 +120t +4t² +900 +300t +25t²

Combine like terms:

t² +4t² +25t² = 30t²

120t +300t = 420t

900 +900 = 1800

So ||v||² = 30t² +420t +1800

To find the minimum, take derivative with respect to t and set to zero.

But since it's a quadratic in t, the minimum occurs at t = -b/(2a) where the quadratic is at² + bt + c.

Here, a = 30, b = 420.

So t = -420/(2*30) = -420/60 = -7

So t = -7

Therefore, the minimal magnitude occurs when t = -7.

Now plug t = -7 into the expressions for v1, v2, v3:

v1 = t = -7

v2 = 30 + 2*(-7) = 30 -14 = 16

v3 = -30 -5*(-7) = -30 +35 = 5

So the vector v is (-7, 16, 5)

Let me check if this satisfies the original cross product equation.

Compute cross product of a = (1,2,-5) and v = (-7,16,5).

Using the cross product formula:

i component: 2*5 - (-5)*16 = 10 +80 =90

j component: - [1*5 - (-5)*(-7)] = - [5 -35] = -(-30) =30

k component: 1*16 -2*(-7) =16 +14=30

So indeed, the cross product is (90,30,30), which matches the given vector. So that's correct.

Therefore, the vector with the smallest magnitude is (-7,16,5). Let me compute its magnitude to check.

||v||² = (-7)^2 +16^2 +5^2 =49 +256 +25=330

So magnitude is sqrt(330). Let me see if that's indeed the minimal.

Alternatively, since we derived it through minimizing the quadratic, which is a parabola opening upwards (since coefficient of t² is positive), so the vertex is indeed the minimum. So yes, that's the minimal magnitude.

Therefore, the answer is (-7,16,5). Wait, but let me just verify once again. Maybe I made a calculation error.

Wait, let's recompute the cross product again step by step.

a × v = determinant of the matrix:

|i   j   k |
|1    2  -5|
|-7 16  5|

So i*(2*5 - (-5)*16) - j*(1*5 - (-5)*(-7)) + k*(1*16 -2*(-7))

Compute each component:

i: 2*5 =10, (-5)*16= -80, so 10 - (-80)=90. Correct.

j: 1*5=5, (-5)*(-7)=35, so 5 -35= -30. Then with the negative sign in front: -(-30)=30. Correct.

k:1*16=16, 2*(-7)=-14, so 16 - (-14)=16+14=30. Correct.

Yes, all components match. So the answer is correct.

So, the vector with the smallest magnitude is (-7,16,5). Therefore, the answer is \boxed{\begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix}}.

**Final Answer**
The vector $\mathbf{v}$ with the smallest magnitude is \boxed{\begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix}}.
Therefore, the vector $\mathbf{v}$ with the smallest magnitude is $\boxed{\begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix}}$.
</think>