Let
A = (0,6) M = (0,3) N = (6,0) C = (12,0)
Slope of AN = ( 6 - 0) / (0 - 6) = -1
Equation of line containing AN ..... y = -1x + 6
Slope of MC = ( 0-3)/(12 - 0) = -3/12 = -1/4
Equation of line containing MC ..... y = (-1/4)x+ 3
x coordinate of P = altitude of triangle AMP
(-1/4)x + 3 = -1x + 6
(3/4) x = 3
x = 4
And [ AMP ] = (1/2) AM * 4 = 2 * 3 = 6
And area of triangle AMC = (1/2) AM BC = (1/2) (3) (12) =18
So area of APC = [ AMC ] - [ AMP] = 18 - 6 = 12