2. Given $-1 \leqslant 2 x+y-z \leqslant 8, 2 \leqslant x-y+z \leqslant 9, -3 \leqslant x+2 y-z \leqslant 7$. Find the maximum and minimum values of the function $f(x, y, z)=7 x+5 y-2 z$.
2. Let $a(2 x+y-z)+b(x-y+z)+c(x+2 y-z)=7 x+5 y-2 z$. By comparing the coefficients on both sides, we get $a=1$, $b=2, c=3$. Multiplying the known three conditions by $1,2,3$ respectively, we get $-1 \leqslant 2 x+y-z \leqslant 8, 4 \leqslant 2(x-y+z) \leqslant 18$, $-9 \leqslant 3(x+2 y-z) \leqslant 21$. Adding these three inequalities, we get $-6 \leqslant 7 x+5 y-2 z \leqslant 47$, hence $-6 \leqslant f(x, y, z) \leqslant 47$, which means the maximum and minimum values of $f(x, y, z)$ are 47 and -6, respectively.