If $P=3 \hat{i}+5 \hat{j}-\hat{k}$ and $Q=\hat{i}+2 \hat{j}+3 \hat{k}$ are two sides of a triangle,then its area is equal to . . . . . . sq units.

If the area of the parallelogram with $\bar{a}$ and $\bar{b}$ as two adjacent sides is $16$ sq. units,then the area of the parallelogram having $3 \bar{a}+2 \bar{b}$ and $\bar{a}+3 \bar{b}$ as two adjacent sides (in sq. units) is

Let $\hat{\alpha}, \hat{\beta}, \hat{\gamma}$ be three unit vectors such that $\hat{\alpha} \times (\hat{\beta} \times \hat{\gamma}) = \frac{1}{2}(\hat{\beta} + \hat{\gamma})$. If $\hat{\beta}$ is not parallel to $\hat{\gamma}$, then the angle between $\hat{\alpha}$ and $\hat{\beta}$ is

Let $ABC$ be a triangle of area $15 \sqrt{2}$ and the vectors $\overrightarrow{AB}=\hat{i}+2 \hat{j}-7 \hat{k}$,$\overrightarrow{BC}=a \hat{i}+b \hat{j}+c \hat{k}$ and $\overrightarrow{AC}=6 \hat{i}+d \hat{j}-2 \hat{k}$,where $d>0$. Then the square of the length of the largest side of the triangle $ABC$ is:

The direction cosines of a line which is perpendicular to lines whose direction ratios are $3, -2, 4$ and $1, 3, -2$ are

Let $\overrightarrow{a}=2 \hat{i}-\hat{j}+3 \hat{k}$,$\overrightarrow{b}=3 \hat{i}-5 \hat{j}+\hat{k}$ and $\vec{c}$ be a vector such that $\vec{a} \times \vec{c}=\vec{c} \times \vec{b}$ and $(\overrightarrow{a}+\overrightarrow{c}) \cdot(\overrightarrow{b}+\overrightarrow{c})=168$. Then the maximum value of $|\vec{c}|^2$ is :