Given Point is P(x,y).

Lines are x + y - 5 = 0 and 3x - 2y + 7 =0

Now distnace of P(x,y) from the line x + y - 5 = 0 is

d1 = (x+y-5)/√(1+1) = (x+y-5)/√2

Again distnace of P(x,y) from the line 3x - 2y + 7 = 0 is

d2 = (3x-2y+7)/√{3² + (-2)² } = (3x-2y+7)/√(9+4) = (3x-2y+7)/√13

Given sum of distances is equal to 10

=> d1 + d2 = 10

=> (x+y-5)/√2 + (3x-2y+7)/√13 = 10

=> √13*(x+y-5) + √2*(3x-2y+7) = 10*√2*√13 

=> x(√13 + 3√2) + y(√13 - 2√2) + (7√2 - 5√13) = 10

=> y(√13 - 2√2) = -(√13 + 3√2)x + 10 - (7√2 - 5√13)

=> y =  {-(√13 + 3√2)/(√13 - 2√2)}x + {10 - (7√2 - 5√13)}/(√13 - 2√2)

This forms a line  y = mx + c

So the given point P(x,y) must move on a line.