Let $P(x, y)$ be a variable point. The perpendicular distance from point $(x, y)$ to line $ax + by + c = 0$ is given by $d = |ax + by + c|/\sqrt{a^2 + b^2}$.

Distance from $P$ to line $x + y - 5 = 0$:

$$d_1 = \frac{|x + y - 5|}{\sqrt{1^2 + 1^2}} = \frac{|x + y - 5|}{\sqrt{2}}$$

Distance from $P$ to line $3x - 2y + 7 = 0$:

$$d_2 = \frac{|3x - 2y + 7|}{\sqrt{3^2 + (-2)^2}} = \frac{|3x - 2y + 7|}{\sqrt{13}}$$

Given: $d_1 + d_2 = 10$

Therefore,

$$\frac{|x + y - 5|}{\sqrt{2}} + \frac{|3x - 2y + 7|}{\sqrt{13}} = 10$$

For the case where $x + y - 5 \geq 0$ and $3x - 2y + 7 \geq 0$:

$$\frac{x + y - 5}{\sqrt{2}} + \frac{3x - 2y + 7}{\sqrt{13}} = 10$$

Multiplying by $\sqrt{26}$:

$$\sqrt{13}(x + y - 5) + \sqrt{2}(3x - 2y + 7) = 10\sqrt{26}$$

Expanding:

$$\sqrt{13}\,x + \sqrt{13}\,y - 5\sqrt{13} + 3\sqrt{2}\,x - 2\sqrt{2}\,y + 7\sqrt{2} = 10\sqrt{26}$$

Rearranging:

$$(\sqrt{13} + 3\sqrt{2})\,x + (\sqrt{13} - 2\sqrt{2})\,y = 10\sqrt{26} + 5\sqrt{13} - 7\sqrt{2}$$

This is a linear equation of the form $Ax + By = C$, representing a straight line.

Similarly, other cases with different sign combinations also yield linear equations. Hence, the locus of point $P$ is a straight line.

∴ $P$ must move on a line.