= Solution
Introduce nonnegative <slack variables> $s_1,s_2,s_3$ and start the <simplex algorithm> at $x=y=z=0$, with slacks $9,8,2$. The objective's largest positive entering coefficient belongs to $y$. Its ratio test is $9/3=3$, $8/2=4$, $2/1=2$, so $s_3$ leaves and $y$ enters.
Solve that pivot row for $y$ and substitute into the other rows and the objective $v$:
$$
\begin{aligned}
y&=2-2x-3z-s_3,\\
s_1&=3-2x-z+3s_3,\\
s_2&=4-x+2z+2s_3,\\
v&=10-7x-15z-5s_3.
\end{aligned}
$$
This is a feasible <simplex dictionary> at nonbasic variables $x=z=s_3=0$. All their objective coefficients are negative, so no feasible increase can improve the objective. Thus
$$
\boxed{(x,y,z)=(0,2,0),\qquad v_{\max}=10.}
$$
The negative reduced coefficients also prove uniqueness of this optimum.
The <dual linear program>, with nonnegative variables $u_1,u_2,u_3$, is
$$
\begin{gathered}
\text{minimize }9u_1+8u_2+2u_3,\\
8u_1+5u_2+2u_3\ge3,\quad3u_1+2u_2+u_3\ge5,\quad10u_1+4u_2+3u_3\ge0.
\end{gathered}
$$
The feasible choice $\boxed{(u_1,u_2,u_3)=(0,0,5)}$ has objective $10$. <Weak duality> proves that it and the primal point are optimal. <Complementary slackness> also forces $u_1=u_2=0$ because the first two primal slacks are positive, and then the positive $y$ forces $u_3=5$.
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