I have down-scaled the problem a little, but the same logic would apply
The image below lays out the particulars :
The base of triangle XYZ = XY = 6 and the height = 6...so...the area =
6^2 / 2 = 18 sq units
Now...X = (12, 12) and Y = (15,6)
And we can call segment XY the base of triangle CXY...and this is
sqrt [ (15 -12)^2 + ( 12 - 6)^2 ] = sqrt [ 3^2 + 6^2] = sqrt (45) = 3sqrt (5) units in length = (1)
And the equation of the line passing through these two points is given by
y = -2(x -12) + 12 ... y = -2x + 36 or 2x + y - 36 = 0 = AX + By + C = 0
Now ....the distance from C to this line is the altitude of CXY and it is given by
altitude = abs (Ax + By + C) / sqrt (A^2 + B^2) where (x,y) are the coordinates of C = (0,0)
So we have that
altitude = abs ( 2(0) + 1(0) - 36 ) / sqrt ( 2^2 + 1^2) = 36 / sqrt (5) = (2)
So.....the area of triangle CXY = base * altitude / 2 = (1) * (2) / 2 =
[ 3 sqrt(5) * 36 ] / [ 2 * sqrt (5) ] = 3 * 36 / 2 = 108 / 2 = 54 sq units
Now.....since in the original problem the area of XYZ is specified as 21 sq units....and the area of XYZ triangle in the diagram = 18 sq units...we can use a proportion to find the true area of CXY....so we have....
21 / 18 = true area of CXY / 54
7 / 6 = true area of CXY / 54
true area of CXY = 7 * 54 / 6 = 7 * 9 = 63 sq units