<Question>
Let $\mathbf{a},$ $\mathbf{b},$ $\mathbf{c}$ be three vectors such that
\[\mathbf{a} \times \mathbf{b} = \begin{pmatrix} 6 \\ -7 \\ 3 \end{pmatrix}, \quad \mathbf{a} \times \mathbf{c} = \begin{pmatrix} 4 \\ 7 \\ 2 \end{pmatrix}, \quad \mathbf{b} \times \mathbf{c} = \begin{pmatrix} 1 \\ -7 \\ 18 \end{pmatrix}.\]Compute $(2 \mathbf{b} - \mathbf{a}) \times (3 \mathbf{c} + \mathbf{a}).$
</Question>

<think>
<Plan_1>
Okay, so I need to compute the cross product of (2b - a) and (3c + a). Let me recall the properties of cross products to see how to approach this. The cross product is linear in both of its arguments, which means that I can distribute the cross product over addition and scalar multiplication. So, expanding this out, I should be able to break it down into several terms involving a x b, a x c, b x c, and maybe some other combinations. Let me write that expansion step by step.

First, let's denote the vectors as given:

<prompt_1.1> Expand the cross product (2b - a) x (3c + a) using the linearity of the cross product. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
a × b = (6, -7, 3)
a × c = (4, 7, 2)
b × c = (1, -7, 18)

We need to compute (2b - a) × (3c + a).

Expanding this using the distributive property:

(2b - a) × (3c + a) = 2b × 3c + 2b × a - a × 3c - a × a

Simplify each term:

First term: 2b × 3c = 6(b × c)
Second term: 2b × a = 2(b × a)
Third term: -a × 3c = -3(a × c)
Fourth term: -a × a = - (a × a)

But remember that the cross product of any vector with itself is zero, so the last term is zero. So we can ignore that.

So now we have:

6(b × c) + 2(b × a) - 3(a × c)
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Now, let's recall that b × a = - (a × b). So, substituting that into the second term:

6(b × c) + 2(- (a × b)) - 3(a × c)
= 6(b × c) - 2(a × b) - 3(a × c)

Now, substitute the given values for a × b, a × c, and b × c.

Given:

<prompt_2.1> Substitute the given cross product values into the expanded expression. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
a × b = (6, -7, 3)
a × c = (4, 7, 2)
b × c = (1, -7, 18)

So substituting:

6*(1, -7, 18) - 2*(6, -7, 3) - 3*(4, 7, 2)
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Let me compute each scalar multiplication first.

<prompt_3.1> Compute each scalar multiplication for the terms 6*(1, -7, 18), -2*(6, -7, 3), and -3*(4, 7, 2). </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
First term: 6*(1, -7, 18) = (6*1, 6*(-7), 6*18) = (6, -42, 108)

Second term: -2*(6, -7, 3) = (-2*6, -2*(-7), -2*3) = (-12, 14, -6)

Third term: -3*(4, 7, 2) = (-3*4, -3*7, -3*2) = (-12, -21, -6)

Now, add these three results together:

(6, -42, 108) + (-12, 14, -6) + (-12, -21, -6)

Let's add the x-components: 6 + (-12) + (-12) = 6 -12 -12 = 6 -24 = -18

Then the y-components: -42 +14 + (-21) = (-42 +14) -21 = (-28) -21 = -49

Then the z-components: 108 + (-6) + (-6) = 108 -6 -6 = 108 -12 = 96

So adding all together gives (-18, -49, 96)
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Wait, let me check the arithmetic again step by step to make sure I didn't make a mistake.

<prompt_4.1> Verify the arithmetic calculations for the cross product. </prompt_4.1>

But let me verify once more. Maybe I made a mistake in signs. Let's check each term again:

<prompt_4.2> Re-verify the cross product calculation for potential sign errors. </prompt_4.2>

But let me check if there's another way to approach this problem or if I can verify the answer through another method. Hmm, cross products can sometimes be verified using properties or by checking if the resulting vector is perpendicular to both (2b -a) and (3c +a). However, without knowing the actual vectors a, b, c, it's hard to verify orthogonality directly.

Alternatively, maybe I can use the determinant formula for cross products, but since we are dealing with a combination of given cross products, perhaps the linearity approach is the most straightforward.

Wait, another way to check: Let's see if the components make sense.

<prompt_4.3> Cross-verify the cross product result for consistency. </prompt_4.3>

Let me just check once more for any possible miscalculations.

First term: 6*(1, -7, 18) = (6, -42, 108). Correct.

Second term: -2*(6, -7, 3) = (-12, 14, -6). Correct.

Third term: -3*(4,7,2) = (-12, -21, -6). Correct.

Adding x-components: 6 -12 -12 = -18. Correct.

<prompt_4.4> Perform a final verification of the cross product calculation. </prompt_4.4>
</Plan_4>

<Execution_4>
<execution_4.1>
First term:

6*(1, -7, 18):

x: 6*1 = 6

y: 6*(-7) = -42

z: 6*18 = 108

Correct.

Second term:

-2*(6, -7, 3):

x: -2*6 = -12

y: -2*(-7) = 14

z: -2*3 = -6

Correct.

Third term:

-3*(4,7,2):

x: -3*4 = -12

y: -3*7 = -21

z: -3*2 = -6

Correct.

Adding x-components:

6 (from first term) + (-12) (from second term) + (-12) (from third term) = 6 -12 -12 = -18

Yes.

Adding y-components:

-42 (from first term) +14 (from second term) + (-21) (from third term) = (-42 +14) = -28; then -28 -21 = -49

Correct.

Adding z-components:

108 (from first term) + (-6) (from second term) + (-6) (from third term) = 108 -6 -6 = 96

Correct.

So the final result is (-18, -49, 96).
</execution_4.1>

<execution_4.2>
First term: 6*(1, -7, 18) is indeed (6, -42, 108). Correct.

Second term: -2*(6, -7, 3). So 6*-2 = -12, -7*-2 = 14, 3*-2 = -6. Correct.

Third term: -3*(4,7,2). 4*-3 = -12, 7*-3 = -21, 2*-3 = -6. Correct.

Adding the components:

x: 6 -12 -12 = 6 -24 = -18. Correct.

y: -42 +14 -21 = (-42 +14) = -28; -28 -21 = -49. Correct.

z: 108 -6 -6 = 108 -12 = 96. Correct.

So the resulting vector is (-18, -49, 96).
</execution_4.2>

<execution_4.3>
Let's see:

First component: 6*1 -12 -12 = 6 -24 = -18. Correct.

Second component: -42 +14 -21 = (-42 -21) +14 = -63 +14 = -49. Correct.

Third component: 108 -6 -6 = 96. Correct.

So the final result is (-18, -49, 96).
</execution_4.3>

<execution_4.4>
y-components: -42 +14 -21 = (-42 +14) = -28; -28 -21 = -49. Correct.

z-components: 108 -6 -6 = 96. Correct.

So, the final result is (-18, -49, 96).
</execution_4.4>
</Execution_4>

<Plan_5>
Based on execution_4.4:
**Final Answer**
\boxed{\begin{pmatrix} -18 \\ -49 \\ 96 \end{pmatrix}}
</Plan_5>

<Execution_5>
No prompt to execute.
</Execution_5>

<Final_answer>
Therefore, the final answer is $\boxed{\begin{pmatrix} -18 \\ -49 \\ 96 \end{pmatrix}}$.
</Final_answer>
</think>