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 $\vec{a} = 3\hat{i} + 2\hat{j} + x\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$,for some real $x$. Then $|\vec{a} \times \vec{b}| = r$ is possible if

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=\hat{i}+\hat{j}-2\hat{k}, \vec{c}=\hat{i}-2\hat{j}+3\hat{k}$ and $\vec{d}=-4\hat{i}+5\hat{j}-3\hat{k}$. If $\vec{d}=x(\vec{b} \times \vec{c})-\frac{7}{9}(\vec{c} \times \vec{a})+z(\vec{a} \times \vec{b})$,then find the value of $x$.

If $a=\hat{i}+\hat{j}$ and $b=3 \hat{i}-2 \hat{j}$,then the vector $r$ satisfying the equations $r \times a=b \times a$ and $r \times b=a \times b$ is

Let $O$ be the origin and the position vector of the point $P$ be $-\hat{i}-2\hat{j}+3\hat{k}$. If the position vectors of the points $A, B$ and $C$ are $-2\hat{i}+\hat{j}-3\hat{k}$,$2\hat{i}+4\hat{j}-2\hat{k}$ and $-4\hat{i}+2\hat{j}-\hat{k}$ respectively,then the projection of the vector $\overline{OP}$ on a vector perpendicular to the vectors $\overline{AB}$ and $\overline{AC}$ is $......$.

Area of a rectangle having vertices $A, B, C$ and $D$ with position vectors $-\hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}, \hat{i}+\frac{1}{2} \hat{j}+4 \hat{k}, \hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}$ and $-\hat{i}-\frac{1}{2} \hat{j}+4 \hat{k}$ respectively is