If $\lambda > 0$,let $\theta$ be the angle between the vectors $\vec{a} = \hat{i} + \lambda \hat{j} - 3 \hat{k}$ and $\vec{b} = 3 \hat{i} - \hat{j} + 2 \hat{k}$. If the vectors $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ are mutually perpendicular,then the value of $(14 \cos \theta)^2$ is equal to

The magnitude of the projection of vector $\vec{a} = -\hat{i} + 2\hat{j} - \hat{k}$ on the vector $\vec{b} = \hat{i} + 2\hat{j} + 2\hat{k}$ is . . . . . . .

Statement $(A):$ If $|\vec{a}| = 2, |\vec{b}| = 3, |2\vec{a} - \vec{b}| = 5$,then $|2\vec{a} + \vec{b}| = 5$.
Reason $(R): |\vec{p} - \vec{q}| = |\vec{p} + \vec{q}|$

Let the vectors $\vec{a} = -\hat{i} + \hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} + 3\hat{j} + \hat{k}$. For some $\lambda, \mu \in \mathbb{R}$, let $\vec{c} = \lambda \vec{a} + \mu \vec{b}$. If $\vec{c} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) = 10$ and $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = -2$, then $|\vec{c}|^2$ is equal to:

If $A(3,4,5), B(4,6,3), C(-1,2,4)$ and $D(1,0,5)$ are such that the angle between the lines $DC$ and $AB$ is $\theta$,then $\cos \theta$ is equal to

Show that the points $A, B$ and $C$ with position vectors $\vec{a}=3 \hat{i}-4 \hat{j}-4 \hat{k}$,$\vec{b}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-3 \hat{j}-5 \hat{k}$ respectively form the vertices of a right-angled triangle.