If $a, b$ and $c$ are three non-collinear points and $ka + 2b + 3c$ is a point in the plane of $a, b$ and $c$, then $k =$

The position vectors of the points $A, B, C$ are $(2i + j - k)$,$(3i - 2j + k)$,and $(i + 4j - 3k)$ respectively. These points

The position vector of the point of intersection of the medians (centroid) of a triangle,whose vertices are $A(1, 2, 3)$,$B(1, 0, 3)$,and $C(4, 1, -3)$ is

In a quadrilateral $PQRS$,$M$ and $N$ are mid-points of the sides $PQ$ and $RS$ respectively. If $\vec{PS} + \vec{QR} = t \vec{MN}$,then $t =$

The value of $k$ for which the vectors $\vec{a} = \hat{i} - \hat{j}$ and $\vec{b} = -2\hat{i} + k\hat{j}$ are collinear is

Given $\overrightarrow{p} = 3 \hat{i} + 2 \hat{j} + 4 \hat{k}$,$\overrightarrow{a} = \hat{i} + \hat{j}$,$\overrightarrow{b} = \hat{j} + \hat{k}$,$\overrightarrow{c} = \hat{i} + \hat{k}$ and $\overrightarrow{p} = x \overrightarrow{a} + y \overrightarrow{b} + z \overrightarrow{c}$,then $x, y, z$ are respectively: