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

Then the value of $x+y-z$ is ( ).
(A) 3
(B) 2
(C) 1
(D) 0
-.1.D.
From the given equation, we have
$$
\left[2(x+2)^{2}+3\right]\left[(y-5)^{2}+4\right]\left[3(z-3)^{2}+5\right] \leqslant 60 \text {. }
$$

Therefore, $x=-2, y=5, z=3$.
Thus, $x+y-z=-2+5-3=0$.