The adjacent sides of a parallelogram are $\vec{a} = 3\hat{i} + \hat{j} + 4\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$. Then,the area of the parallelogram is . . . . . . sq. units.

Find $|\vec{a} \times \vec{b}|,$ if $\vec{a}=\hat{i}-7 \hat{j}+7 \hat{k}$ and $\vec{b}=3 \hat{i}-2 \hat{j}+2 \hat{k}.$

The area of the triangle with vertices $(1,2,0)$,$(1,0,2)$ and $(0,3,1)$ is.

The vectors $a = xi + yj + zk$ and $b = j$ are such that $a, c, b$ form a right-handed system. Then $c$ is

If $\vec{a}=2 \hat{i}-\hat{j}+3 \hat{k}, \vec{b}=-3 \hat{i}+5 \hat{j}-4 \hat{k}$ and $\vec{c}=6 \hat{i}-4 \hat{j}+5 \hat{k}$,then $(\vec{a} \times \vec{b}) \cdot(\vec{b} \times \vec{c})=$

Consider $P(1, 2, -3)$,$Q(-2, 1, -4)$,and $R(3, 4, -2)$. Let $\vec{B} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$,where $A_x, A_y$,and $A_z$ are the projections of the area of triangle $PQR$ on the $yz, zx$,and $xy$ planes,respectively. Then,the value of $|\vec{B}|^2$ is: