If $2x+3y=7$ and $x–y=1$ are two normals of a parabola $y^2=4ax$ from a point then third normal may be

Clearly point of intersection of normals is (2,1) and as sum of slopes is zero
$\Rightarrow m_1+m_2+m_3=0$
$\Rightarrow -\frac{2}{3}+1+m_3=0$
$\Rightarrow m_3=-\frac{1}{3}$
$\therefore$ equation is  $y-1=-\frac{1}{3}(x-2)$
$\Rightarrow x+3y=5$