Let A be the point of intersection of the lines 3 x + 2 y = 14 , 5 x - y = 6 and B be the point of…

Solving lines $L_1(3x+2y=14)$ and $L_2(5x-y=6)$.

$\Rightarrow x=\frac{14-2y}{3}=\frac{6+y}{5}$

$\Rightarrow 70-10y=18+3y$

$\Rightarrow 52=13y$

$\Rightarrow y=4$

$\Rightarrow x=\frac{14-8}{3}=2$

$\Rightarrow A\equiv \left(2,4\right)$

Now, solving lines $L_3(4x+3y=8)$ and $L_4(6x+y=5)$

$\Rightarrow x=\frac{8-3y}{4}=\frac{5-y}{6}$

$\Rightarrow 48-18y=20-4y$

$\Rightarrow 28=14y$

$\Rightarrow y=2$

$\Rightarrow x=\frac{8-6}{4}=\frac{1}{2}$

$\Rightarrow B\equiv \left(\frac{1}{2},2\right)$

So, equation of $AB$ is given by,

$AB: \left(y-4\right)=\left(\frac{2-4}{{\displaystyle \frac{1}{2}-2}}\right)\left(x-2\right)$

$\Rightarrow AB: \left(y-4\right)=\left(\frac{4}{{\displaystyle 3}}\right)\left(x-2\right)$

$\Rightarrow AB: \left(y-4\right)=\left(\frac{4}{{\displaystyle 3}}\right)\left(x-2\right)$

$\Rightarrow 3y-12=4x-8$

$\Rightarrow 4x-3y+4=0$

The distance of $P\left(5,-2\right)$ from the line $AB$ is given by,

$D=\left|\frac{4\left(5\right)-3(-2)+4}{5}\right|$

$\Rightarrow D=6$ units.