From ∣z−6∣=5, point z lies on circle centered at (6,0) with radius 5.
From ∣z+2−6i∣=5, point lies on circle centered at (−2,6) with radius 5.
Let z=x+iy.
First condition gives: x2+y2=12x−11.
Second gives: x2+y2=−4x+12y−15.
Setting equal: 16x−12y+4=0, so x=43y−1.
Substituting into (x−6)2+y2=25: (3y−25)2+16y2=400, which simplifies to 25y2−150y+225=0, giving y=3 and x=2.
Thus z=2+3i, so z2=−5+12i and z3=−46+9i.
z3+3z2−15z+141=(−46+9i)+3(−5+12i)−15(2+3i)+141=50