8. Given real numbers $x, y, z$ satisfy
$$
\begin{array}{l}
\left(2 x^{2}+8 x+11\right)\left(y^{2}-10 y+29\right)\left(3 z^{2}-18 z+32\right) \\
\leqslant 60 .
\end{array}
$$

Then $x+y-z=$ $\qquad$
8. 0 .

Original expression
$$
\begin{array}{l}
=\left[2(x+2)^{2}+3\right]\left[(y-5)^{2}+4\right]\left[3(z-3)^{2}+5\right] \\
\leqslant 60 \\
\Rightarrow x=-2, y=5, z=3 \\
\Rightarrow x+y-z=0 .
\end{array}
$$