Explanation

(c)
Let $z = x + iy$, then $\overline{z} = x - iy$
$\therefore\ \ \ \ z + \overline{z} = 2x$ and $z - \overline{z} = 2iy$
Given, $(3 + i)\left( z + \overline{z} \right) - (2 + i)\left( z - \overline{z} \right) + 14i = 0$
$\Rightarrow \ \ \ (3 + i)2x - (2 + i)2iy + 14i = 0$
$\Rightarrow \ \ \ 6x + 2ix - 4yi + 2y + 14i = 0 + oi$
On comparing real and imaginary part, we get
$6x + 2y = 0$
And $2x - 4y + 14 = 0$
On solving, we get $x = - 1,\ y = 3$
$\therefore\ \ \ \ z\overline{z} = |z|^{2} = \left( \sqrt{( - 1)^{2} + (3)^{2}} \right)^{2} = 10$