Let $x+y=k$ be normal to
$y^{2}=12 x$ at $P(\alpha, \beta)$...(i)
$\therefore \beta^{2}=12 \alpha$...(ii)
Also, slope of normal at $P(\alpha, \beta)$ is -1
Form, Eq.(i),
$\frac{d y}{d x} =\frac{6}{y}$
$\Rightarrow \left(\frac{d y}{d x}\right)_{P} =\frac{6}{\beta}$
$\therefore -1 =\frac{-1}{6 / \beta}$
$\Rightarrow \beta=6, \alpha=3$
$\therefore P$ is (3,6) which lies on $x+y=k$
$\therefore 3+6=k$
$\Rightarrow k=9$