= Solution
For any feasible $x,y$, increasing $z$ lowers the objective, so the additional upper bound makes $z^*=0$. Feasibility then requires $2x-y\leq2$. The lowest feasible value of $y$ on the circle occurs at the lower intersection with $2x-y=2$. Solving gives the candidate
$$
x^*=\frac{4-\sqrt{21}}5,\qquad y^*=-\frac{2+2\sqrt{21}}5,\qquad z^*=0.
$$
A global certificate avoids relying on the circle sketch. Introduce <Lagrange multipliers> for $g\leq0$, $z\leq0$ and $h=0$:
$$
L=3y-z+\lambda g+\nu z+\mu h.
$$
Choose
$$
\mu=\frac3{\sqrt{21}}>0,\qquad \lambda=\frac{3(\sqrt{21}-4)}{5\sqrt{21}}>0,\qquad \nu=1+\lambda>0.
$$
The coefficient of $z$ vanishes, and $2\mu x^*+2\lambda=0$, $2\mu y^*+3-\lambda=0$. Therefore <completing the square> yields
$$
L=3y^*+\mu\bigl((x-x^*)^2+(y-y^*)^2\bigr).
$$
The candidate minimizes $L$ globally and satisfies both active inequality constraints, so <complementary slackness> and the <Lagrangian sufficiency theorem> give
$$
\boxed{\min(3y-z)=-\frac{6(1+\sqrt{21})}{5},\quad (x^*,y^*,z^*)=\left(\frac{4-\sqrt{21}}5,-\frac{2+2\sqrt{21}}5,0\right).}
$$
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