If the area of the parallelogram with $\bar{a}$ and $\bar{b}$ as two adjacent sides is $16$ sq. units,then the area of the parallelogram having $3 \bar{a}+2 \bar{b}$ and $\bar{a}+3 \bar{b}$ as two adjacent sides (in sq. units) is

If $a = 2i + k$,$b = i + j + k$ and $c = 4i - 3j + 7k$. If $d \times b = c \times b$ and $d \cdot a = 0$,then $d$ 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:

Let the vectors $\vec{a}, \vec{b}, \vec{c}$ be given as $\vec{a} = a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}$,$\vec{b} = b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}$,and $\vec{c} = c_{1} \hat{i}+c_{2} \hat{j}+c_{3} \hat{k}$. Then show that $\vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}$.

Suppose $L_1$ and $L_2$ are two lines having the direction ratios $1, -2, -2$ and $0, 2, 1$ respectively. If the direction cosines of a line perpendicular to both $L_1$ and $L_2$ are $l, m, n$,then $|l| + |m| + |n| =$

If $a=\hat{i}+\hat{j}+\hat{k}$,$c=\hat{j}-\hat{k}$,$a \times b=c$ and $a \cdot b=3$,then $b$ is equal to