Given $P Q R S$ a quadritateral $A$ divides $S R$ in the ratio $1: 3$ and $B$ is the mid-point $P R$.

Let the position vector of $P, Q, R, S, A, B$ are $\mathbf{p}, \mathbf{q}, \mathbf{r}_r$ $\mathbf{s}$, $\mathbf{a}$ and $\mathbf{b}$ respectively
$\begin{aligned}
\therefore \quad \mathbf{S R} & =\mathbf{r}-\mathbf{s} ; \mathbf{Q R}=\mathbf{r}-\mathbf{q} \\
\mathbf{P S} & =\mathbf{s}-\mathbf{p} ; \mathbf{P Q}=\mathbf{q}-\mathbf{p} \\
\mathbf{A B} & =\mathbf{b}-\mathbf{a} \\
\mathbf{a} & =\frac{3 \mathbf{s}+\mathbf{r}}{4} ; \mathbf{b}=\frac{\mathbf{p}+\mathbf{r}}{2}
\end{aligned}$
Now, 3SE- QR - 3PS - PQ $=\mathbf{K A B}$
$\begin{aligned}
& \begin{aligned}
& 3(\mathbf{r}-\mathbf{s})-(\mathbf{r}-\mathbf{q})-3(\mathbf{s}-\mathbf{p})-(\mathbf{q}-\mathbf{p}) \\
&=k(\mathbf{b}-\mathbf{a})
\end{aligned} \\
& \begin{array}{l}
\Rightarrow 3 \mathbf{r}-3 \mathbf{s}-\mathbf{r}+\mathbf{q}-3 \mathbf{s}+3 \mathbf{p}-\mathbf{q}+\mathbf{p}=k \\
\left(\frac{\mathbf{p}+\mathbf{r}}{2}-\frac{3 \mathbf{s}+\mathbf{r}}{4}\right)
\end{array} \\
& \quad=2 \mathbf{r}-6 \mathbf{s}+4 \mathbf{p}=k\left(\frac{2 \mathbf{p}+2 \mathbf{r}-3 \mathbf{s}-\mathbf{r}}{4}\right) \\
& \quad=8 \mathbf{r}-24 \mathbf{s}+16 \mathbf{p}=2 k \mathbf{p}+k \mathbf{r}-3 k \mathbf{s} \\
& \therefore \quad k=8
\end{aligned}$