Okay, so I need to compute the cross product of (2b - a) and (3c + a). Let me recall how cross products work. I remember that cross products are distributive over vector addition, so I can expand this expression using the distributive property. Then, maybe I can use the given cross products to substitute values. Let me try to write this out step by step.

First, let's expand (2b - a) × (3c + a). Using the distributive property of cross products, this should be equal to 2b × 3c + 2b × a - a × 3c - a × a. Let me check that. Yes, each term in the first vector is dotted with each term in the second vector, so that's correct.

But wait, cross product is anti-commutative, right? So a × b = - (b × a). So when I have terms like -a × 3c, that's the same as -3(a × c). Similarly, -a × a is zero because the cross product of any vector with itself is zero. So let me rewrite each term:

2b × 3c = 6(b × c)
2b × a = 2(b × a)
- a × 3c = -3(a × c)
- a × a = 0

So combining these, the entire expression becomes 6(b × c) + 2(b × a) - 3(a × c). 

Now, let's look at the given cross products. We know a × b, a × c, and b × c. But in our expression, we have b × a, a × c, and b × c. Since a × b is given as (6, -7, 3), then b × a would be the negative of that, right? Because a × b = - (b × a). So b × a = - (6, -7, 3) = (-6, 7, -3).

Similarly, a × c is given as (4, 7, 2), so that term is already provided. And b × c is given as (1, -7, 18). So we can substitute these values into our expression.

Let me write each term with their coefficients:

First term: 6(b × c) = 6*(1, -7, 18) = (6*1, 6*(-7), 6*18) = (6, -42, 108)

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

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

Now, sum these three vectors component-wise:

First components: 6 + (-12) + (-12) = 6 -12 -12 = 6 -24 = -18

Second components: -42 +14 + (-21) = -42 +14 -21 = (-42 +14) = -28; -28 -21 = -49

Third components: 108 + (-6) + (-6) = 108 -6 -6 = 108 -12 = 96

So the resulting vector is (-18, -49, 96). Let me check my calculations again to make sure I didn't make any arithmetic errors.

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 components:

First components: 6 -12 -12 = 6 -24 = -18. Correct.

Second components: -42 +14 = -28; -28 -21 = -49. Correct.

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

Hmm, seems like all steps check out. So the final answer should be (-18, -49, 96). Let me just verify once more if I substituted the cross products correctly.

Original expression: (2b -a) × (3c +a) = 2b×3c +2b×a -a×3c -a×a.

Yes, correct. Then substituting:

2*3 =6, so 6(b×c) =6*(1,-7,18) = (6,-42,108)

2b×a =2*(-6,7,-3)= (-12,14,-6)

-3a×c = -3*(4,7,2)= (-12,-21,-6)

Adding them up gives (-18,-49,96). That seems right. So I think that's the answer.

**Final Answer**
The result of $(2 \mathbf{b} - \mathbf{a}) \times (3 \mathbf{c} + \mathbf{a})$ is $\boxed{\begin{pmatrix} -18 \\ -49 \\ 96 \end{pmatrix}}$.
Therefore, the cross product of $(2 \mathbf{b} - \mathbf{a})$ and $(3 \mathbf{c} + \mathbf{a})$ is $\boxed{\begin{pmatrix} -18 \\ -49 \\ 96 \end{pmatrix}}$.
</think>