Step 1:
Tangent to $y^2=16x$ perpendicular to $3x-y+8=0$
Compare with $y^2=4ax,4a=16\Rightarrow a=4$
Any tangent to $y^2=4ax$ is $y=mx+\large\frac{a}{m}$
Here $y=mx+\large\frac{4}{m}$
Step 2:
Required tangent is to $3x-y+8=0$ whose slope is $3$.
Therefore slope of the tangent =$\large\frac{-1}{3}$$=m$
The required tangent is $y=-\large\frac{1}{3}$$x+\large\frac{4}{\Large\frac{-1}{3}}$
$\Rightarrow y=\large\frac{-x}{3}$$-12$
$\Rightarrow x+3y+36=0$