jmc

Sum $3a-1$, $a^2+1$, and $a^2+2$ to find $2a^2+3a+2=16$. Subtract 16 from both sides and factor the left-hand side to find $(2a+7)(a-2)=0⟹a=-7/2$ or $a=2$. Discarding the negative solution, we substitute $a=2$ into $3a-1$, $a^2+1$, and $a^2+2$ to find that the side lengths of the triangle are 5, 5, and 6 units. Draw a perpendicular from the 6-unit side to the opposite vertex to divide the triangle into two congruent right triangles (see figure). The height of the triangle is $\sqrt{5^2-3^2}=4$ units, so the area of the triangle is $\frac{1}{2}(6)(4)=\boxed{12\text{ square units}}$.