= Solution
With $g=2x-y-z-2$ and $h=x^2+y^2-5$, the maximization <optimization Lagrangian> is
$$
L=-2\lambda x+(3+\lambda)y+(\lambda-1)z-\mu(x^2+y^2-5)+2\lambda.
$$
A finite unconstrained maximum requires $\lambda=1$ to cancel the coefficient of $z$. With $\mu=1$, <completing the square> gives
$$
L=12-(x+1)^2-(y-2)^2.
$$
Thus its global maximizers have $x=-1$ and $y=2$. The equality constraint holds, and <complementary slackness> with $\lambda=1$ makes $g=0$, giving $z=-6$. The <Lagrangian sufficiency theorem> certifies
$$
\boxed{(x^*,y^*,z^*)=(-1,2,-6),\qquad \max(3y-z)=12.}
$$
Every feasible point has objective at most $L\leq12$, so this is a global conclusion rather than just a stationary-point calculation.
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