Discussion
If \(\vec{a} \times (\hat{i} - \hat{j} + \hat{k})\) = \((\hat{i} - \hat{j} + \hat{k}) \times \vec{b}\) and \(|\vec{a} + \vec{b}| = 3\sqrt{3}\) , then the possible values of \((\vec{a} + \vec{b}).(3\hat{i} + 2\hat{j} + \hat{k})\) are
Answer: D
The possible values are \(\pm 6\).
Let \(\vec{v} = \hat{i} - \hat{j} + \hat{k}\). Since \(\vec{v} \times \vec{b} = -\vec{b} \times \vec{v}\), the condition becomes \((\vec{a} + \vec{b}) \times \vec{v} = \vec{0}\), so \(\vec{a} + \vec{b}\) is parallel to \(\vec{v}\).
\[ \begin{aligned} \vec{a} + \vec{b} &= \lambda(\hat{i} - \hat{j} + \hat{k}) \\ |\lambda|\sqrt{3} &= 3\sqrt{3} \Rightarrow \lambda = \pm 3 \\ (\vec{a} + \vec{b})\cdot(3, 2, 1) &= \lambda(3 - 2 + 1) = \pm 6 \end{aligned} \]
Forgetting that \(\lambda\) can be negative would give only +6, and using \(\lambda = \pm 1\) gives \(\pm 2\).
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