If $a$ is collinear with $b = 3 \hat{i} + 6 \hat{j} + 6 \hat{k}$ and $a \cdot b = 27$,then $|a| =$

If the position vectors of the vertices of a triangle are $6i + 4j + 5k$,$4i + 5j + 6k$,and $5i + 6j + 4k$,then the triangle is

Statement $(A):$ If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a} + \vec{b} + \vec{c} = 0$,then $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{3}{2}$.
Reason $(R): (\vec{x} + \vec{y})^2 = |\vec{x}|^2 + |\vec{y}|^2 + 2(\vec{x} \cdot \vec{y})$.

$OABCD$ is a pentagon in which the sides $OA$ and $CB$ are parallel and the sides $OD$ and $AB$ are parallel. Also,it is given that $\frac{OA}{CB}=2$,$\frac{OD}{AB}=\frac{1}{3}$. If $\vec{OA}=\vec{a}, \vec{OD}=\vec{d}$,then $\vec{AD}+\vec{OC}+\vec{DC}=$

The direction cosines of the vector $\vec{a} = -2 \hat{i} + \hat{j} - 5 \hat{k}$ are

The position vectors of the points $P$ and $Q$ are respectively $-2 \bar{i}-3 \bar{j}+\bar{k}$ and $3 \bar{i}+3 \bar{j}+2 \bar{k}$. The ratio in which the point having position vector $\frac{-9}{2} \bar{i}-6 \bar{j}+\frac{1}{2} \bar{k}$ divides the line segment joining $P$ and $Q$ is