Explanation

(c)
The combined equation of the pair of tangents drawn from $(1,\ 2)$ to the ellipse ${3x}^{2} + {2y}^{2} = 5$ is $SS^{'} = T^{2}$
$\Rightarrow \left( {3x}^{2} + {2y}^{2} - 5 \right)\left\lbrack {3(1)}^{2} + {2(2)}^{2} - 5 \right\rbrack = \left\lbrack 3x(1) + 2y(2) - 5 \right\rbrack^{2}$
$\Rightarrow \left( {3x}^{2} + {2y}^{2} - 5 \right)(3 + 8 - 5) = (3x + 4y - 5)^{2}$
$\Rightarrow 9x^{2} - 24xy - 4y^{2} + 40y + 30x - 55 = 0$
This is the equation of pair of straight lines,
Where, $a = 9,\ h = - 12,\ b = - 4$
The angle between these lines is given by
$\tan\theta = \frac{2\sqrt{h^{2} - ab}}{a + b}$
$\Rightarrow \tan\theta = \frac{2\sqrt{144 + 36}}{9 - 4} = \frac{2\sqrt{180}}{5} = \frac{12}{\sqrt{5}}$
$\Rightarrow \theta = \tan^{- 1}\left( \frac{12}{\sqrt{5}} \right)$