= Solution
The amended payoff is $P_{\mathrm{new}}=P+3x_3$. Using the previous <simplex dictionary>,
$$
P_{\mathrm{new}}=8+\tfrac12x_3-\tfrac43z_1-\tfrac16z_2.
$$
Thus $x_3$ enters. The positive pivot-column coefficients give ratios $3/(1/2)=6$ from $x_2$ and $(29/2)/10=29/20$ from $z_3$; the coefficient $-3/2$ in the $x_1$ row gives no bound. The <simplex ratio test> makes $z_3$ leave. Dividing that row by $10$ and eliminating $x_3$ gives
$$
\begin{aligned}
x_1+\tfrac{19}{120}z_1+\tfrac{17}{120}z_2+\tfrac3{20}z_3&=\tfrac{107}{40},\\
x_2+\tfrac9{40}z_1+\tfrac7{40}z_2-\tfrac1{20}z_3&=\tfrac{91}{40},\\
x_3+\tfrac{13}{60}z_1-\tfrac1{60}z_2+\tfrac1{10}z_3&=\tfrac{29}{20},\\
P_{\mathrm{new}}+\tfrac{173}{120}z_1+\tfrac{19}{120}z_2+\tfrac1{20}z_3&=\tfrac{349}{40}.
\end{aligned}
$$
The positive objective-row coefficients and nonnegative <slack variables> prove optimality at $z_1=z_2=z_3=0$. Therefore
$$
\boxed{x=(\tfrac{107}{40},\tfrac{91}{40},\tfrac{29}{20}),\qquad\min(2x_1-3x_2-5x_3)=-\tfrac{349}{40}.}
$$
All three constraints are equalities at this point. Strict positivity of the three objective-row coefficients also forces every optimal point to have all three slacks zero; the displayed dictionary then makes the optimizer unique.
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