If $a, b, c$ are the position vectors of three collinear points,then the existence of scalars $x, y, z$ (not all zero) is such that:

If the vectors $\hat{i}+2 \hat{j}+x \hat{k}$ and $y \hat{i}+6 \hat{j}+4 \hat{k}$ are collinear,then the values of $x$ and $y$ are respectively,

Let $\overrightarrow{OA} = \hat{i} - 3\hat{j} + \hat{k}$, $\overrightarrow{OB} = \hat{i} + 3\hat{j} - 2\hat{k}$, and $\overrightarrow{OC} = 4\hat{i} + 3\hat{j} + 5\hat{k}$ be the position vectors of three points $A$, $B$, and $C$. Let $P$ be the point which divides $AB$ in the ratio $2:1$. If $l, m, n$ are the direction cosines of the vector $\overrightarrow{PC}$, then $l + 3m + 2n =$

In $\triangle ABC$,if $D$ and $E$ are the mid-points of the sides $BC$ and $CA$ respectively,then $2(\vec{AD}+\vec{EB})=$

If $\overline{a} = m \overline{b} + n \overline{c}$,where $\overline{a} = 4 \hat{i} + 13 \hat{j} - 18 \hat{k}$,$\overline{b} = \hat{i} - 2 \hat{j} + 3 \hat{k}$,and $\overline{c} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$,then $m + n =$

If $a = (2, 5)$ and $b = (1, 4),$ then the vector parallel to $(a + b)$ is