Explanation

(d)
Equation of given ellipse is $\frac{{(x - 1)}^{2}}{9} + \frac{{(y - 2)}^{2}}{4} = 1$
Equation of normal ellipse $\frac{X^{2}}{9} + \frac{Y^{2}}{4} = 1$ is
$3X\ \sec\theta - 2Y\ cosec\ \theta = 5$
Slope of normal is $\frac{3}{2}\tan\theta$
Which is parallel to $3x - y = 1,$ then $\frac{3}{2}\tan{\theta = 3}$
$\Rightarrow \tan\theta = 2$
$\therefore\sin\theta = \frac{2}{\sqrt{5}},\ \cos\theta = \frac{1}{\sqrt{5}}$
So, equation of normal is $3\sqrt{5}X - \sqrt{5}Y = 5$
$\because X = x - 1,\ Y = y - 2$
$\therefore 3\sqrt{5}(x - 1) - \sqrt{5}(y - 2) = 5$
$\Rightarrow \sqrt{5}(3x - y) = 5(\sqrt{5} + 1)$
$\Rightarrow 3x - y = \sqrt{5}(\sqrt{5} + 1)$