= Solution
The feasible point $x=(8/5,0,1/5)$ makes the second and third constraints tight; the first has left side two. Its objective is $27/5$.
To prove optimality from first principles, multiply each of the second and third inequalities by $3/5$ and add. For every nonnegative feasible $x$,
$$
3x_1+x_2+3x_3\le3x_1+\frac{12}5x_2+3x_3\le\frac35(5+4)=\frac{27}5.
$$
This is an explicit <weak duality> bound, proved here simply by adding inequalities. The displayed point attains it, so
$$
\boxed{\phi(0)=27/5,\qquad x^*=(8/5,0,1/5).}
$$
Equality forces $x_2=0$ and both contributing constraints tight, proving uniqueness as well.
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