If the collinear points $A, B$ and $C$ have position vectors respectively $(1, x, 3), (3, 4, 7)$ and $(y, -2, -5)$, then $x+y=$

Let $O$ be the origin, $\vec{OP} = \vec{a}$ and $\vec{OQ} = \vec{b}$. If $R$ is the point on $\vec{OP}$ such that $\vec{OP} = 5\vec{OR}$, and $M$ is the point such that $\vec{OQ} = 5\vec{RM}$, then $\vec{PM}$ is equal to:

Let $\vec{a}=x \hat{i}+y \hat{j}+z \hat{k}$ and $x=2 y$. If $|\vec{a}|=5 \sqrt{2}$ and $\vec{a}$ makes an angle of $135^{\circ}$ with the $z$-axis,then $\vec{a}=$

If $a, b, c$ are the position vectors of the vertices $A, B, C$ of the triangle $ABC,$ then the centroid of $\Delta ABC$ is

If $\vec{a} = \hat{i} + \hat{j} - 2 \hat{k}$,$\vec{b} = 2 \hat{i} - \hat{j} + \hat{k}$ and $\vec{c} = 3 \hat{i} - \hat{k}$. If $\vec{c} = m \vec{a} + n \vec{b}$ then $m + n = $

The diagonals of a parallelogram are the vectors $\vec{d_1} = 3 \hat{i} + 6 \hat{j} - 2 \hat{k}$ and $\vec{d_2} = -\hat{i} - 2 \hat{j} - 8 \hat{k}$. Then the length of the shorter side of the parallelogram is