Given vertices of quadrilateral $\mathrm{ABCD}$ are $\mathrm{B}(2,3)$ $\mathrm{C}(5,-2)$ and $\mathrm{D}(1,-1)$ with area of $\mathrm{ABCD}$ is 10 sq. units.
Area of quadrilateral $\mathrm{ABCD}=$
$$
\pm \frac{1}{2}\left\{\left(\mathrm{x}_1-\mathrm{x}_3\right)\left(\mathrm{y}_2-\mathrm{y}_1\right)-\left(\mathrm{x}_2-\mathrm{x}_4\right)\left(\mathrm{y}_1-\mathrm{y}_3\right)\right\}
$$
Let the vertex $A$ is $(x, y)$
Area of quadrilateral $\mathrm{ABCD}=$
$$
\begin{aligned}
& \pm \frac{1}{2}\{(\mathrm{x}-5)(3+1)-(2-1)(\mathrm{y}+2)\} \\
& 10=\frac{1}{2}\{4(\mathrm{x}-5)-(\mathrm{y}+2)\} \\
& 4 \mathrm{x}-20-\mathrm{y}-2=20 \\
& 4 \mathrm{x}-\mathrm{y}-22=20 \\
& 4 \mathrm{x}-\mathrm{y}-42=0
\end{aligned}
$$
Also, $10=-\frac{1}{2}\{4(\mathrm{x}-5)-(\mathrm{y}+2)\}$
$$
\begin{aligned}
& 20=-\{4 x-20-y-2\} \\
& 20=-\{4 x-y-22\}
\end{aligned}
$$
$$
\begin{aligned}
& 4 x-y-22=-20 \\
& 4 x-y-2=0
\end{aligned}
$$
Locus of point $A$ is $(4 x-y-42)(4 x-y-2)=0$ So,option (d) is correct.