= Solution
The same <weak duality> certificate gives the upper bound $(27+3\epsilon_2+3\epsilon_3)/5$ for every feasible point. Keep $x_2=0$ and solve the tight second and third constraints:
$$
x_1=\frac{16+4\epsilon_2-\epsilon_3}{10},\qquad x_3=\frac{2-2\epsilon_2+3\epsilon_3}{10}.
$$
These are nonnegative precisely when their numerators are nonnegative. The remaining first-constraint slack is $4+\epsilon_1-\epsilon_3/2$. All three are positive for sufficiently small perturbations, so the point is feasible and attains the bound. This illustrates <linear programming sensitivity within a fixed optimal basis>:
$$
\boxed{\phi(\epsilon)=\frac{27+3\epsilon_2+3\epsilon_3}5\quad\text{near }0.}
$$
More generally this expression is valid throughout the region specified by those three feasibility inequalities.
With $\epsilon_1=\epsilon_2=0$, the conditions reduce to
$$
\boxed{-\frac23\le\epsilon_3\le8.}
$$
The endpoints are included. Outside this interval the bound cannot be attained: its equality conditions require exactly the point above, which then has a negative $x_3$ or violates the first constraint. Whenever feasible, the problem attains a maximum because the first constraint and nonnegativity bound all coordinates, so its value is strictly smaller outside the interval. For $\epsilon_3<-4$ it is infeasible. Thus the range is exact, not just a sufficient neighborhood.
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