Solution:

The five unit squares have total area $5,$ so each region must have area $\dfrac52.$

The region to the lower right of the line is a right triangle with legs $3-a$ and $3,$ minus the one unit square it does not cover. Setting its area to $\dfrac52$ gives $$\dfrac{3(3-a)}{2}-1=\dfrac52,$$ so $3(3-a)=7$ and $a=\dfrac23.$

Thus, the correct answer is C.