Given pair of lines are
$x^{2}-3xy+2y^{2}=0$
and $x^{2}-3xy+2y^{2}+x-2=0$
$\therefore \left(x-2y\right)\left(x-y\right)=0$
and $\left(x-2y+2\right)\left(x-y-1\right)=0$
$\Rightarrow x-2y=0, x-y=0$ and
$x-2y+2=0, x-y-1=0$
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.