If $a=\hat{i}+2 \hat{j}+3 \hat{k}$,$b=-\hat{i}+2 \hat{j}+\hat{k}$,$c=\hat{i}+2 \hat{j}-2 \hat{k}$,$n$ is perpendicular to both $a$ and $b$,and $\theta$ is the angle between $c$ and $n$,then $\sin \theta=$

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 $(\vec{a} \times \vec{b}) \cdot(\vec{b} \times \vec{c})=$

If $\overline{OA} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\overline{OB} = 4\hat{i} + \hat{k}$ are the position vectors of the points $A$ and $B$, then the position vector of a point on the line passing through $B$ and parallel to the vector $\overline{OA} \times \overline{OB}$ which is at a distance of $\sqrt{189}$ units from $B$ is

For $\vec{a}$ and $\vec{b}$,$|\vec{a}|=3$,$|\vec{b}|=\frac{\sqrt{2}}{3}$ and $\vec{a} \times \vec{b}$ is a unit vector,then the angle between $\vec{a}$ and $\vec{b}$ is . . . . . . .

Using vectors,find the area of the triangle $ABC$ with vertices $A(1, 2, 3)$,$B(2, -1, 4)$ and $C(4, 5, -1)$.

If the area of the parallelogram with $\vec{a}$ and $\vec{b}$ as two adjacent sides is $15$ sq. units,then the area of the parallelogram having $3 \vec{a}+\vec{b}$ and $\vec{a}+3 \vec{b}$ as two adjacent sides,in square units,is