Area of Triangle Formed by a Point and the Centre

Normal at $$P$$.
$$\frac{3x}{\cos\theta} - \frac{2y}{\sin\theta} = 5$$

$$G$$ on $$x$$-axis ($$y=0$$):
$$x = \frac{5\cos\theta}{3} \implies G = \left(\frac{5\cos\theta}{3},\ 0\right)$$
$$H$$ on $$y$$-axis ($$x=0$$):
$$-\frac{2y}{\sin\theta} = 5 \implies y = -\frac{5\sin\theta}{2} \implies H = \left(0,\ -\frac{5\sin\theta}{2}\right)$$
Area of triangle $$OGH$$.
$$\text{Area} = \frac{1}{2} \cdot |OG| \cdot |OH| = \frac{1}{2} \cdot \frac{5\cos\theta}{3} \cdot \frac{5\sin\theta}{2}$$
$$= \frac{25\sin\theta\cos\theta}{12} = \frac{25\sin 2\theta}{24}$$

$$\sin 2\theta$$ is maximum when $$2\theta = \dfrac{\pi}{2}$$, i.e.:
$$\theta = \frac{\pi}{4}$$
$$\text{Maximum Area} = \frac{25}{24}$$
$$\boxed{(C) \quad \theta = \frac{\pi}{4}}$$