Let $P(6 \cos \theta, 3 \sqrt{3} \sin \theta)$ be any point on the ellipse $\frac{x^2}{36}+\frac{y^2}{27}=1$. The equation of the tangent at $P(6 \cos \theta, 3 \sqrt{3} \sin \theta)$ is $\frac{x}{6} \cos \theta+\frac{y}{3 \sqrt{3}} \sin \theta=1$.
$$
\Rightarrow \quad 3 \sqrt{3} x \cos \theta+6 y \sin \theta-18 \sqrt{3}=0
$$
The product of the lengths of the perpendiculars from $(3,0)$ and $(-3,0)$ an Eq. (i) is given by
$$
\begin{aligned}
& P=\left|\frac{3 \times 3 \sqrt{3} \cos \theta-18 \sqrt{3}}{\sqrt{27 \cos ^2 \theta+36 \sin ^2 \theta}}\right|\left|\frac{3 \times 3 \sqrt{3} \cos \theta+18 \sqrt{3}}{\sqrt{27 \cos ^2 \theta+36 \sin ^2 \theta}}\right| \\
& =\frac{36 \times 27-9 \times 27 \cos ^2 \theta}{36 \sin ^2 \theta+27 \cos ^2 \theta} \\
& =\frac{9 \times 27\left(4-\cos ^2 \theta\right)}{36\left(1-\cos ^2 \theta\right)+27 \cos ^2 \theta} \\
& =\frac{9 \times 27\left(4-\cos ^2 \theta\right)}{36-9 \cos ^2 \theta} \\
& =\frac{9 \times 27\left(4-\cos ^2 \theta\right)}{9\left(4-\cos ^2 \theta\right)}=27 . \\
&
\end{aligned}
$$