The cosine of the angle $A$ of the triangle with vertices $A(1, -1, 2)$,$B(6, 11, 2)$,and $C(1, 2, 6)$ is:

$A$ force of magnitude $5$ units acting along the vector $2i - 2j + k$ displaces the point of application from $(1, 2, 3)$ to $(5, 3, 7)$. The work done is:

$|a \times b|^2 + (a \cdot b)^2 = ?$

The orthogonal projection of vector $a$ on vector $b$ is given by:

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=3 \hat{i}+2 \hat{j}-\hat{k}, \vec{c}=\lambda \hat{j}+\mu \hat{k}$ and $\hat{d}$ be a unit vector such that $\vec{a} \times \hat{d}=\vec{b} \times \hat{d}$ and $\vec{c} \cdot \hat{d}=1$. If $\vec{c}$ is perpendicular to $\vec{a}$,then $|3 \lambda \hat{d}+\mu \vec{c}|^2$ is equal to . . . . . . .

If $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$,$|\vec{a}|=3$,$|\vec{b}|=5$,and $|\vec{c}|=7$,then the angle between $\vec{a}$ and $\vec{b}$ is