Let a circle $\mathrm{C}$ of radius 5 lie below the $\mathrm{x}$-axis. The line $L_1=4x+3y-2$ passes through the centre $P$ of the circle $\mathrm{C}$ and intersects the line ${\mathrm{L}}_2:3\mathrm{x}-4\mathrm{y}-11=0$ at $\mathrm{Q}$. The line ${\mathrm{L}}_2$ touches $\mathrm{C}$ at the point $\mathrm{Q}$. Then the distance of $\mathrm{P}$ from the line $5x-12y+51=0$ is

$L_1:4x+3y+2=0$

$$L_2:3x-4y-11=0$$

Since circle $C$ touches the line $L_2$ at $Q$ intersection point $Q$ of $L_1$ and $L_2$, is $(1,-2)$
$\because P$ lies of $L_1$

$$\therefore P(x,-\frac{1}{3}(2+4x))$$

Now, $PQ=5\Rightarrow (x-1{)}^2+{(\frac{4x+2}{3}-2)}^2=25$

$$\begin{array}{rl} & \Rightarrow (x-1{)}^2[1+\frac{16}{9}]=25\\ & \Rightarrow (x-1{)}^2=9\\ & \Rightarrow x=4,-2\end{array}$$
$\because$ Circle lies below the $x$-axis

$$\begin{array}{rl}\therefore & y=-6\\ & P(4,-6)\end{array}$$

Now distance of $P$ from $5x-12y+51=0$

$$=\left|\frac{20+72+51}{13}\right|=\frac{143}{13}=11$$