Explanation

(c)
Let any point $P(x_{1},y_{1})$ outside the circle. Then, equation of tangent to the circle $x^{2} + y^{2} + 6x + 6y = 2$ at the point $P$ is
$xx_{1} + yy_{1}3\left( x + x_{1} \right) + 3\left( y + y_{1} \right) - 2 = 0$ …(i)
The Eq. (i) and the line $5x - 2y + 6 = 0$ intersect at a point $Q$ on $y$-axis $ie,\ x = 0$
$\Rightarrow 5(0) - 2y + 6 = 0\ \ \Rightarrow y = 3$
$\therefore$ Coordinates of $Q$ are (0, 3)
Point $Q$ satisfies Eq. (i)
$\therefore 3x_{1} + 6y_{1} + 7 = 0$ …(ii)
Distance between $P$ and $Q$ is given by
$PQ^{2} = x_{1}^{2} + \left( y_{1} - 3 \right)^{2}$
$= x_{1}^{2} + y_{1}^{2} - 6y_{1} + 9$
$= 11 - 6x_{1} - 12y_{1}\ \ \left( \because\ \ x_{1}^{2} + y_{1}^{2} + 6x_{1} + 6y_{1} - 2 = 0 \right)$
$= 11 - 2(3x_{1} - 6y_{1})$
$= 11 - 2( - 7) = 25$ [from Eq. (ii)]
$\therefore PQ = 5$