The tangent at $(1,7)$ to the parabola $x^{2}=y-6$ is
$x =\frac{1}{2}(y+7)-6$
$\Rightarrow 2 x =y+7-12$
$\Rightarrow y =2 x+5$
which is also a tangent to the circle
$x^{2}+y^{2}+16 x+12 y +c=0$
$\therefore x^{2}+(2 x+5)^{2}+ 16 x+12(2 x+5)+c=0$
$\Rightarrow 5 x^{2}+60 x+85+c=0$
must have equal roots.
Let $\alpha$ and $\beta$ be the roots of the equation.
Then, $\alpha+\beta=-12 \Rightarrow \alpha=-6 (\because \alpha=\beta)$
$\therefore x=-6$ and $y=2 x+5=-7$
$\Rightarrow $ Point of contact is $(-6,-7)$