$\hat{i} \cdot (\hat{j} \times \hat{k}) + \hat{j} \cdot (\hat{i} \times \hat{k}) + \hat{k} \cdot (\hat{i} \times \hat{j}) + \hat{j} \cdot (\hat{j} \times \hat{k}) = $ . . . . . . .

For three unit vectors $\vec{a}, \vec{b}, \vec{c}$ satisfying $|\vec{a}-\vec{b}|^{2}+|\vec{b}-\vec{c}|^{2}+|\vec{c}-\vec{a}|^{2}=9$ and $|2\vec{a}+k\vec{b}+k\vec{c}|=3$, the positive value of $k$ is:

If the line joining points $A$ and $B$ having position vectors $6 \vec{a}-4 \vec{b}+4 \vec{c}$ and $-4 \vec{c}$ respectively,and the line joining the points $C$ and $D$ having position vectors $-\vec{a}-2 \vec{b}-3 \vec{c}$ and $\vec{a}+2 \vec{b}-5 \vec{c}$ intersect,then their point of intersection is

Let $\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k}, \vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k}$ and $\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}$. If a vector $\vec{d}$ satisfies $\vec{d} \times \vec{b}=\vec{c} \times \vec{b}$ and $\vec{d} \cdot \vec{a}=24$,then $|\vec{d}|^2$ is equal to $.........$.

If $A = (k, 1, -1)$,$B = (2k, 0, 2)$,and $C = (2 + 2k, k, 1)$,and $AB \perp BC$,then the value of $k$ is:

If $\overrightarrow{F_1} = i - j + k,$ $\overrightarrow{F_2} = -i + 2j - k,$ $\overrightarrow{F_3} = j - k,$ $\vec{A} = 4i - 3j - 2k$ and $\vec{B} = 6i + j - 3k,$ then the scalar product of $(\overrightarrow{F_1} + \overrightarrow{F_2} + \overrightarrow{F_3})$ and $\overrightarrow{AB}$ will be: