Explanation

(d)
These lines cannot be the sides of a rectangle as none of these are parallel nor they are perpendicular.
$Now,\ for\ concurrent\ \left| \begin{matrix}
1 & 2 & - 3 \\
3 & 4 & - 7 \\
2 & 3 & - 4
\end{matrix} \right|$
$= 1( - 16 + 21) - 2(2) - 3(1)$
$\neq 0$
Hence, these are not concurrent.
Opposite side of the parallelogram are
$x^{2} - 5x + 6 = 0$ and $y^{2} - 6y + 5 = 0$
$\Rightarrow (x - 2)(x - 3) = 0\$and $(y - 1)(y - 5) = 0$
$\Rightarrow x - 2 = 0,\ x - 3 = 0\$and $y - 1 = 0,\ y - 5 = 0$
$\therefore\$Vertices are $(3,\ 5),\ (2,5),\ (2,1)$ and (3,1)