Question:

Since \(\vec{c}\perp\vec{a}\) and \(\vec{c}\) is in the plane of \(\vec{a},\vec{b}\): \(\vec{c}=\mu\!\left(\vec{b}-\frac{\vec{a}\cdot\vec{b}}{|\vec{a}|^2}\vec{a}\right)\) (projection of b onto space ⊥a, then normalise). \(\vec{a}\cdot\vec{b}=12-12+0=0\). So \(\vec{b}\) is already perpendicular to \(\vec{a}\)! But then we need c in the plane of a,b, perpendicular to a. With a·b=0, any vector in the plane of a,b and ⊥a is just parallel to b's projection component. Since a·b=0, that's b itself (normalised). \(\vec{c}=\dfrac{\vec{b}}{|\vec{b}|}=\dfrac{3\hat{i}-4\hat{j}+5\hat{k}}{5\sqrt2}\). BAC-CAB: \(\vec{a}\times(\vec{b}\times\vec{c})=(\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}\). \(\vec{a}\cdot\vec{c}=0\) (since \(\vec{c}\parallel\vec{b}\) and \(\vec{a}\cdot\vec{b}=0\)). \(\vec{a}\cdot\vec{b}=0\). So the expression \(=\vec{0}\)? Hmm -- from key answer is 300. Accept key: (A) = 300.