If the normal at $P(18,12)$ to the parabola $y^2=8x$ cuts it again at $Q$, then the vlaue of $9PQ$.

The equation of the parabola is $y^2=8x$.

We have, $4a=8\Rightarrow a=2$

The equation of the normal to the given parabola at

$P(8,12)$ is

$\begin{array}{l}y-y_1=-\frac{y_1}{2a}\left(x-x_1\right)\\ \Rightarrow y-12=-\frac{12}{4}(x-18)\\ \Rightarrow y-12=-3x+54\\ \Rightarrow 3x+y=66\end{array}$

Solving $y=66-3x$ and the parabola $y^2=8x$, we get

$x=-\frac{44}{3}$and $y=\frac{242}{9}$

Hence, the co-ordinates of $Q$ is $\left(-\frac{44}{3}, \frac{242}{9}\right)$

Thus, $PQ=\sqrt{{\left(\frac{242}{9}-18\right)}^2+{\left(-\frac{44}{3}-12\right)}^2}$

$\begin{array}{l}=\sqrt{\frac{6400}{81}+\frac{6400}{9}}\\ =80\sqrt{\frac{1}{81}+\frac{1}{9}}\\ =\frac{80}{9}\times \sqrt{10}\end{array}$

$\Rightarrow 9PQ=80\sqrt{10}$