If $a \times (b \times c) = 0,$ then

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} = 0$ and the angle between $\vec{b}$ and $\vec{c}$ is $\pi / 3$,then $\vec{a}$ is equal to

Let $\vec{a} = 4\hat{i} - \hat{j} + 3\hat{k}$, $\vec{b} = 10\hat{i} + 2\hat{j} - \hat{k}$ and a vector $\vec{c}$ be such that $2(\vec{a} \times \vec{c}) + 3(\vec{b} \times \vec{c}) = \vec{0}$. If $\vec{a} \cdot \vec{c} = 15$, then $\vec{c} \cdot (\hat{i} + \hat{j} - 3\hat{k})$ is equal to:

$\vec{a}=\hat{i}+\hat{j}-2 \hat{k}$, $\vec{b}=\hat{i}-2 \hat{j}+\hat{k}$ and $\vec{c}=2 \hat{i}+\hat{j}-\hat{k}$ are three vectors. If $\vec{d}$ is a normal to the plane of $\vec{a}$ and $\vec{b}$ and $\vec{d} \cdot \vec{c}=2$, then $|\vec{d}|=$

If $\theta$ is the angle between $\vec{f}=\hat{i}+2 \hat{j}-3 \hat{k}$ and $\vec{g}=2 \hat{i}-3 \hat{j}+a \hat{k}$ and $\sin \theta=\sqrt{\frac{24}{28}}$,then $7 a^2+24 a=$

If $A, B, C, D$ are four points in space,then $|\overline{AB} \times \overline{CD} + \overline{BC} \times \overline{AD} + \overline{CA} \times \overline{BD}| = \lambda \times (\text{Area of } \Delta ABC)$. Find $\lambda$.