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 $a=$

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

$A (2,6,2), B (-4,0, \lambda), C (2,3,-1)$ and $D (4,5,0)$,with $|\lambda| \leq 5$,are the vertices of a quadrilateral $ABCD$. If its area is $18$ square units,then $5-6 \lambda$ is equal to $.........$.

The vector of magnitude $6$ units and perpendicular to vectors $2 \hat{i}+\hat{j}-3 \hat{k}$ and $\hat{i}-2 \hat{j}+\hat{k}$ is

If $1, 2, 3$ and $-1, 0, 1$ are the direction ratios of the rays $OA$ and $OB$ respectively, then the direction cosines of a normal to the plane $AOB$ are

If the vectors $a$ and $b$ are mutually perpendicular,then $a \times \{ a \times \{ a \times (a \times b)\} \}$ is equal to