Write the relation as $(x-8)^2 + (y-8)^2 + (z-8)^2 = 192$ and recall that a square gives the remainder $0$ or $1$ when divided at $4$ to infer that $x-8$, $y-8$, $z-8$ are all even. Set $x-8 = 2a_1$, $y-8 = 2b_1$, $z-8 = 2c_1$ to get $a_1^2 + b_1^2 + c_1^2 = 48$. Repeat the above argument to write $a_1 = 2a_2$, $b_1 = 2b_2$, $c_1 = 2c_2$ with $a_2^2 + b_2^2 + c_2^2 = 12$, then $a_2 = 2a_3$, $b_2 = 2b_3$, $c_2 = 2c_3$ with $a_3^2 + b_3^2 + c_3^2 = 3$.

The latter equality holds for $a_3, b_3, c_3 = \pm 1$, that is whenever $x-8$, $y-8$ and $z-8$ are equal to $8$ or $-8$. Consequently, the triples are $(0, 0, 0)$, $(0, 0, 16)$, $(0, 16, 0)$, $(16, 0, 0)$, $(0, 16, 16)$, $(16, 0, 16)$, $(16, 16, 0)$ and $(16, 16, 16)$.