The area of the parallelogram with vertices $A(1, 2, 3)$,$B(1, 3, a)$,$C(3, 8, 6)$,and $D(3, 7, 3)$ is $\sqrt{265}$ sq. units. Then $a=$

Let $\vec{a}=\hat{i}-2\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$ be two vectors. If $\vec{c}$ is a vector such that $\vec{b} \times \vec{c}=\vec{b} \times \vec{a}$ and $\vec{c} \cdot \vec{a}=0,$ then $\vec{c} \cdot \vec{b}$ is equal to

The adjacent sides of a parallelogram are $\bar{a} = \hat{j} + 2\hat{k}$ and $\bar{b} = \hat{i} + 2\hat{j}$. Find its area.

Let $A(2, 3, 0)$, $B(0, 3, 2)$ and $C(4, 0, 3)$ be the vertices of a triangle. Find the area of the triangle.

$A$ force $\overrightarrow{F} = 2i + j - k$ acts at a point $A$,whose position vector is $2i - j$. The moment of $\overrightarrow{F}$ about the origin is:

The locus of the point $P(\vec{r})$ which forms a triangle $ABP$ of area $1$ sq. unit with the fixed points $A(\hat{i})$ and $B(\hat{j})$ is