Explanation

(d)
The length of tangent from the point (1, 2) to the circle $x^{2} + y^{2} + x + y - 4 = 0$ is $\sqrt{1 + 4 + 1 + 2 - 4},\ \ \ ie,\ 2$
And the length of tangent from the point (1, 2) to the circle
$3x^{2} + 3y^{2} - x - y - k = 0$ is $\sqrt{3 + 12 - 1 - 2 - k}\ \ ie,\ \ \sqrt{12 - k}$
$\therefore\ \ \frac{2}{\sqrt{12 - k}} = \frac{4}{3}$
$\Rightarrow \ \ \frac{3}{2} = \sqrt{12 - k}$
$\Rightarrow \frac{9}{4} = 12 - k\ \ \Rightarrow \ \ k = \frac{39}{4}$