Let $$A$$ be the point of intersection of the lines $$3x + 2y = 14, 5x - y = 6$$ and $$B$$ be the point of intersection of the lines $$4x + 3y = 8, 6x + y = 5$$. The distance of the point $$P(5, -2)$$ from the line $$AB$$ is

We need to find the distance of point $$P(5, -2)$$ from line $$AB$$, where $$A$$ and $$B$$ are intersection points of given pairs of lines.

To begin, we determine point A, the intersection of $$3x + 2y = 14$$ and $$5x - y = 6$$. From the second equation we have $$y = 5x - 6$$, and substituting this into the first equation gives $$3x + 2(5x - 6) = 14$$, hence $$3x + 10x - 12 = 14$$, which simplifies to $$13x = 26$$ and so $$x = 2$$. Substituting back yields $$y = 5(2) - 6 = 4$$, and therefore $$A = (2, 4)$$.

Similarly, point B is the intersection of $$4x + 3y = 8$$ and $$6x + y = 5$$. From the second equation we get $$y = 5 - 6x$$, and substitution into the first yields $$4x + 3(5 - 6x) = 8$$, so $$4x + 15 - 18x = 8$$, then $$-14x = -7$$ giving $$x = \frac{1}{2}$$. It follows that $$y = 5 - 6 \times \frac{1}{2} = 2$$, hence $$B = \left(\frac{1}{2}, 2\right)$$.

Next we find the equation of line AB by computing its slope $$m = \frac{4 - 2}{2 - \frac{1}{2}} = \frac{2}{\frac{3}{2}} = \frac{4}{3}$$ and using the point-slope form with point $$A(2, 4)$$: $$y - 4 = \frac{4}{3}(x - 2)$$. Multiplying through yields $$3(y - 4) = 4(x - 2)$$, which simplifies to $$3y - 12 = 4x - 8$$ and hence to the standard form $$4x - 3y + 4 = 0$$.

Finally, we use the distance formula $$d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$$ for point $$P(5, -2)$$ and line $$4x - 3y + 4 = 0$$: $$d = \frac{|4(5) - 3(-2) + 4|}{\sqrt{16 + 9}} = \frac{|20 + 6 + 4|}{\sqrt{25}} = \frac{30}{5} = 6$$.

Answer: Option 4 — $$6$$