Explanation

(a)
The plane is perpendicular to the line $\frac{x - a}{\cos\theta} = \frac{y + 2}{\sin\theta} = \frac{\mathcal{z} - 3}{0}$
Hence, the direction ratios of the normal of the plane are $\cos\theta,\sin\theta$, and 0 (i)
Now, the required plane passes through the $z$-axis. Hence the point (0, 0, 0) lies on the plane
From Eqs. (i) and (ii), we get equation of the plane as
$\cos\theta(x - 0) + \sin\theta(y - 0) + 0(z - 0) = 0$
$\cos\theta x + \sin\theta y = 0$
$x + y\tan\theta = 0$