(A) 193

(B) 205

(C) 218

(D) 233

Correct Answer: 218

Explanation

Given,

$$ \vec a = 2\hat i - 5\hat j + 5\hat k,\quad \vec b = \hat i - \hat j + 3\hat k $$

From

$$ 2(\vec a \times \vec c) + 3(\vec b \times \vec c) = \vec 0 $$

Using distributive property of cross product,

$$ (2\vec a + 3\vec b)\times \vec c = \vec 0 $$

This implies \(\vec c\) is parallel to \(2\vec a + 3\vec b\).

Compute:

$$ 2\vec a + 3\vec b = 2(2,-5,5) + 3(1,-1,3) $$ $$ = (4,-10,10) + (3,-3,9) = (7,-13,19) $$

So let

$$ \vec c = \lambda(7,-13,19) $$

Now,

$$ \vec a - \vec b = (2,-5,5) - (1,-1,3) = (1,-4,2) $$

Given:

$$ (\vec a - \vec b)\cdot \vec c = -97 $$ $$ (1,-4,2)\cdot \lambda(7,-13,19) = -97 $$ $$ \lambda(7 + 52 + 38) = -97 $$ $$ 97\lambda = -97 \Rightarrow \lambda = -1 $$

Thus,

$$ \vec c = (-7,13,-19) $$

Now,

$$ \vec c \times \hat k = \begin{vmatrix} \hat i & \hat j & \hat k \\ -7 & 13 & -19 \\ 0 & 0 & 1 \end{vmatrix} $$ $$ = 13\hat i + 7\hat j $$

Hence,

$$ |\vec c \times \hat k|^2 = 13^2 + 7^2 = 169 + 49 = 218 $$

Therefore, the correct answer is 218.

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