= Solution
Use payoff $P=-(2x_1-3x_2-2x_3)$, so maximization of $P$ is equivalent to the required minimization. In the given <simplex tableau>, $x_1$ can increase the payoff. The <simplex ratio test> has candidate ratios $3/6=1/2$ in the $z_2$ row and $15/1=15$ in the $z_3$ row; the negative coefficient in the $x_2$ row imposes no upper bound. Thus $x_1$ enters the <simplex basis> and $z_2$ leaves.
Divide the pivot row by $6$, then eliminate $x_1$ from the other rows. The resulting <simplex dictionary> is
$$
\begin{aligned}
x_1-\tfrac32x_3-\tfrac16z_1+\tfrac16z_2&=\tfrac12,\\
x_2+\tfrac12x_3+\tfrac13z_1+\tfrac16z_2&=3,\\
z_3+10x_3+\tfrac{13}6z_1-\tfrac16z_2&=\tfrac{29}2,\\
P+\tfrac52x_3+\tfrac43z_1+\tfrac16z_2&=8.
\end{aligned}
$$
All <slack variables> and original variables are nonnegative. The last row therefore proves $P\le8$ for every <feasible point>; setting the nonbasic variables $x_3,z_1,z_2$ to zero attains it and leaves all basic variables nonnegative. Hence
$$
\boxed{(x_1,x_2,x_3)=(\tfrac12,3,0),\qquad\min(2x_1-3x_2-2x_3)=-8.}
$$
The slack vector is $(0,0,29/2)$.
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