For $\triangle A B C$, we have sides are
$2 x^2-y^2=0, x+y-1=0$
$(\sqrt{2} x-y)(\sqrt{2} x+y)=0, x+y-1=0$
$\sqrt{2} x-y=0, \sqrt{2} x+y=0, x+y-1=0$
On solving these, we get vertices
$(0,0),(\sqrt{2}-1,2-\sqrt{2}),(-\sqrt{2}-1,2+\sqrt{2})$
For $\triangle P Q R$, we have sides are
$2 x^2-5 x y+2 y^2=0,7 x-2 y-12=0$
$(2 x-y)(x-2 y)=0,7 x-2 y-12=0$
$2 x-y=0, x-2 y=0,7 x-2 y-12=0$
On solving these, we get vertices $(0,0),(4,8),(2,1)$
Now, centroid of $\triangle A B C$
$G=\left(\frac{0+\sqrt{2}-1-\sqrt{2}-1}{3}, \frac{0+2-\sqrt{2}+2+\sqrt{2}}{3}\right)$
$=\left(\frac{-2}{3}, \frac{4}{3}\right)$
Orthocentre of $\triangle P Q R$,
$H=\left(\frac{(8-1)(8+8)}{16-4}, \frac{(2-4)(8+8)}{16-4}\right)=\left(\frac{28}{3},-\frac{8}{3}\right)$
$\therefore$ Required distance=G H
$=\sqrt{\left(\frac{28}{3}+\frac{2}{3}\right)^2+\left(\frac{-8}{3}-\frac{4}{3}\right)^2}$
$=\sqrt{100+16}=\sqrt{116}=2 \sqrt{29}$