Step 1:
Equation of the tangent and normal to $y^2=12x$ at $(3,-6)$
The equation of the tangent is given by $yy_1=12\large\frac{(x+x_1)}{2}$
Where $(x_1,y_1)=(3,-6)$
(i.e) $y(-6)=6(x+3)$
$\Rightarrow 6x+6y+18=0$
$\Rightarrow x+y+3=0$
Step 2:
The normal is $\perp$ to the tangent.So its equation is of the form $x-y+k=0$.It passes through $(3,-6)$
Therefore $3+6+k=0$
$\Rightarrow k==-9$
The normal is $x-y-9=0$
Step 3:
The equation of tangent : $x+y+3=0$
The equation of normal : $x-y-9=0$