Let $$\vec{a} = 2\hat{i} + 7\hat{j} - \hat{k}$$, $$\vec{b} = 3\hat{i} + 5\hat{k}$$ and $$\vec{c} = \hat{i} - \hat{j} + 2\hat{k}$$. Let $$\vec{d}$$ be a vector which is perpendicular to both $$\vec{a}$$ and $$\vec{b}$$, and $$\vec{c} \cdot \vec{d} = 12$$. Then $$(-\hat{i} + \hat{j} - \hat{k}) \cdot (\vec{c} \times \vec{d})$$ is equal to

We are given $$\vec{a} = 2\hat{i} + 7\hat{j} - \hat{k}$$, $$\vec{b} = 3\hat{i} + 5\hat{k}$$, and $$\vec{c} = \hat{i} - \hat{j} + 2\hat{k}$$. We seek a vector $$\vec{d}$$ that is perpendicular to both $$\vec{a}$$ and $$\vec{b}$$ while satisfying $$\vec{c}\cdot\vec{d} = 12$$, and then we will compute $$(-\hat{i} + \hat{j} - \hat{k}) \cdot (\vec{c} \times \vec{d})$$.

Because $$\vec{d}$$ is perpendicular to both given vectors, it must be a scalar multiple of their cross product: $$\vec{d} = \lambda \, (\vec{a} \times \vec{b})$$ for some scalar $$\lambda$$. Calculating this cross product via the determinant
$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 7 & -1 \\ 3 & 0 & 5 \end{vmatrix} = \hat{i}(35 - 0) - \hat{j}(10 + 3) + \hat{k}(0 - 21) = 35\hat{i} - 13\hat{j} - 21\hat{k}\,. $$

We then impose the condition $$\vec{c} \cdot \vec{d} = 12$$, which gives
$$\vec{c} \cdot \vec{d} = \lambda\,\vec{c} \cdot (\vec{a} \times \vec{b}) = \lambda\bigl(1\cdot35 + (-1)(-13) + 2(-21)\bigr) = \lambda(35 + 13 - 42) = 6\lambda = 12\,, $$
and hence $$\lambda = 2$$. Substituting back yields $$\vec{d} = 2(35\hat{i} - 13\hat{j} - 21\hat{k}) = 70\hat{i} - 26\hat{j} - 42\hat{k}\,.$$

Next, to find $$\vec{c} \times \vec{d}$$ we evaluate
$$\vec{c} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 70 & -26 & -42 \end{vmatrix} = \hat{i}(42 + 52) - \hat{j}(-42 - 140) + \hat{k}(-26 + 70) = 94\hat{i} + 182\hat{j} + 44\hat{k}\,. $$

Finally, taking the dot product with $$-\hat{i} + \hat{j} - \hat{k}$$ gives
$$(-\hat{i} + \hat{j} - \hat{k}) \cdot (94\hat{i} + 182\hat{j} + 44\hat{k}) = -94 + 182 - 44 = 44\,. $$

The answer is Option B: 44.