Explanation

(c)
Given pair of lines are
$x^{2} - 3xy + {2y}^{2} = 0$
and $x^{2} - 3xy + {2y}^{2} + x - 2 = 0$
$\therefore\ \ (x - 2y)(x - y) = 0$
and $(x - 2y + 2)(x - y - 1) = 0$
$\Rightarrow x - 2y = 0,\ x - y = 0$ and $x - 2y + 2 = 0,\ x - y - 1 = 0$
Since, the lines $x - 2y = 0,\ x - 2y + 2 = 0\ and\ x - y = 0\ ,\ x - y - 1 = 0$ are parallel.
Also, angle between $x - 2y = 0$ and $x - y = 0$ is not $90{^\circ}$
$\therefore$ It is a parallelogram.