Let $z$ be a complex number such that $|z-6|=5$ and $|z+2-6 i|=5$. Then the value of $z^{3}+3 z^{2}-15 z+141$ is equal to

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: $x^2 + y^2 = 12x - 11$.
Second gives: $x^2 + y^2 = -4x + 12y - 15$.
Setting equal: $16x - 12y + 4 = 0$, so $x = \frac{3y - 1}{4}$.
Substituting into $(x-6)^2 + y^2 = 25$: $(3y - 25)^2 + 16y^2 = 400$, which simplifies to $25y^2 - 150y + 225 = 0$, giving $y = 3$ and $x = 2$.
Thus $z = 2 + 3i$, so $z^2 = -5 + 12i$ and $z^3 = -46 + 9i$.
$z^3 + 3z^2 - 15z + 141 = (-46 + 9i) + 3(-5 + 12i) - 15(2 + 3i) + 141 = 50$