Explanation

(b)
Equation of tangent at $(3\sqrt{3},\cos\theta,\sin\theta)$ is
$\frac{x\cos\theta}{3\sqrt{3}} + \frac{y\sin\theta}{1} = 1$
Thus, sum of intercepts$= \left( 3\sqrt{3}\sec\theta + cosec\ \theta \right) = f(\theta)$ [say]
$\Rightarrow \ \ \ f^{'}(\theta) = \frac{3\sqrt{3}\sin^{3}{\theta - \cos^{3}\theta}}{\sin^{2}{\theta\cos^{2}\theta}}$
Put $f^{'}(\theta) = 0$
$\therefore\ \ \sin^{3}{\theta = \frac{1}{3^{3/2}}}\cos^{3}\theta$
$\Rightarrow \ \ \tan{\theta = \frac{1}{\sqrt{3}}\ \ \ \ \Rightarrow \ \ \ \ \theta = \frac{\pi}{6}}$
Also, for $0 < \theta < \frac{\pi}{6},\frac{dz}{d\theta} < 0\ \$ and for $\frac{\pi}{6} < \theta < \frac{\pi}{2},\ \frac{dz}{d\theta} > 0$
$\therefore$Minimum at$\ \theta = \frac{\pi}{6}$