Given circle, $x^2+y^2+6 x+4 y-12=0$

Here, centre is $O(-3,-2)$
and radius $=\sqrt{9+4+12}=\sqrt{25}=5$


Shortest distance of the point $P(2,-14)$ to the circle is
$$
\text { circle is } \begin{aligned}
A P & =d \\
A P & =O P-O A \\
(O A & =R) \\
O P & =\sqrt{(-3-2)^2+(-2+14)^2} \\
& =\sqrt{25+144}=\sqrt{169}=13 \\
\Rightarrow \quad A P & =13-5=8=d
\end{aligned}
$$

Length of tangent from the point $P(2,-14)$ is
$$
\begin{aligned}
& =\sqrt{(2)^2+(-14)^2+6 \cdot 2+4(-14)-12} \\
& =\sqrt{4+196+12-56-12} \\
& =\sqrt{144}=12=1
\end{aligned}
$$

So, required $\sqrt{d+l}=\sqrt{8+12}=\sqrt{20}=2 \sqrt{5}$.