Example 11 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 $u=7 x+5 y-2 z$.
Solution: 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 given three conditions by $1, 2, 3$ respectively, we obtain $-1 \leqslant 2 x+y-z \leqslant 8, 4 \leqslant$ $2(x-y+z) \leqslant 18, -9 \leqslant 3(x+2 y-z) \leqslant 21$. Adding these three inequalities, we get
$$
-6 \leqslant 7 x+5 y-2 z \leqslant 47,
$$

which means $-6 \leqslant u \leqslant 47$.
Therefore, the maximum and minimum values of $u$ are 47 and -6, respectively.