Answer
$$3$$
Work Step by Step
Given $$y=8-3 x, \quad y=6-x, \quad y=2$$
we find the intersection points
$$8-3x = 6-x\to x=1$$
and $y=5$; we solve with respect to $y$, since $$6-y\geq \frac{1}{3}(8-y) \ \ \text{ for }\ \ 2\leq y\leq 5 $$
Hence
\begin{align*}
A&= \int_{2}^{5}[(6-y)-\frac{1}{3}(8-y)]dy\\
&=\left(6y-\frac{1}{2}y^2\right)-\frac{1}{3}\left(8y-\frac{1}{2}y^2\right)\bigg|_{2}^{5}\\
&=3
\end{align*}
We use the Mathematica program to plot the region.
