If the sum of the perpendicular distances of a variable point $P(x, y)$ from the lines $x + y - 5 = 0$ and $3x - 2y + 7 = 0$ is always $10$,show that $P$ must move on a line.

(N/A) The equations of the given lines are:
$x + y - 5 = 0$ $(1)$
$3x - 2y + 7 = 0$ $(2)$
The perpendicular distances of $P(x, y)$ from lines $(1)$ and $(2)$ are given by:
$d_{1} = \frac{|x + y - 5|}{\sqrt{1^{2} + 1^{2}}} = \frac{|x + y - 5|}{\sqrt{2}}$
$d_{2} = \frac{|3x - 2y + 7|}{\sqrt{3^{2} + (-2)^{2}}} = \frac{|3x - 2y + 7|}{\sqrt{13}}$
Given that $d_{1} + d_{2} = 10$,we have:
$\frac{|x + y - 5|}{\sqrt{2}} + \frac{|3x - 2y + 7|}{\sqrt{13}} = 10$
Assuming the expressions inside the modulus are positive,we get:
$\sqrt{13}(x + y - 5) + \sqrt{2}(3x - 2y + 7) = 10\sqrt{26}$
Expanding the terms:
$(\sqrt{13} + 3\sqrt{2})x + (\sqrt{13} - 2\sqrt{2})y - (5\sqrt{13} - 7\sqrt{2} + 10\sqrt{26}) = 0$
This is of the form $Ax + By + C = 0$,which represents a straight line. Since the signs of the expressions inside the modulus can change,$P$ moves on one of the four possible lines formed by the combinations of signs. Thus,$P$ moves on a line.