Let A be the point of intersection of the lines 3x + 2y = 14, 5x -...

Solving lines ${L}_{1} ( 3x + 2y = 14 )$ and ${L}_{2} ( 5x - y = 6 )$.

$⇒ x = \dfrac{14 - 2y}{3} = \dfrac{6 + y}{5}$

$⇒ 70 - 10y = 18 + 3y$

$⇒ 52 = 13y$

$⇒ y = 4$

$⇒ x = \dfrac{14 - 8}{3} = 2$

$⇒ A ≡ \left(2, 4\right)$

Now, solving lines ${L}_{3} ( 4x + 3y = 8 )$ and ${L}_{4} ( 6x + y = 5 )$

$⇒ x = \dfrac{8 - 3y}{4} = \dfrac{5 - y}{6}$

$⇒ 48 - 18y = 20 - 4y$

$⇒ 28 = 14y$

$⇒ y = 2$

$⇒ x = \dfrac{8 - 6}{4} = \dfrac{1}{2}$

$⇒ B ≡ \left(\dfrac{1}{2}, 2\right)$

So, equation of $AB$ is given by,

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

$⇒ AB: \left(y - 4\right) = \left(\dfrac{4}{3}\right)\left(x - 2\right)$

$⇒ AB: \left(y - 4\right) = \left(\dfrac{4}{3}\right)\left(x - 2\right)$

$⇒ 3y - 12 = 4x - 8$

$⇒ 4x - 3y + 4 = 0$

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

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

$⇒ D = 6$ units.