Note that these equation cannot have two common roots. Because if they did, then one equation is a scalar multiple of the other, in which case, $$\dfrac{2}{2}=\dfrac{-3}{-(a+3)}=\dfrac{1}{3a}$$ The first equality would mean that, $3=a+3 \implies a=0$ which makes the second equality absurd.
So, let $\beta$ be the common root of these two equations. We have,
$$2\beta^2-3\beta+1=0$$ $$2\beta^2-(a+3)\beta+3a=0$$ Now we use the Cramer's rule to observe that,
We have that, $$ \dfrac{\beta^2}{\left| \begin{array}{rr} -3 & 1 \\ -(a+3) & 3a \end{array} \right| }=\dfrac{\beta}{\left| \begin{array}{rr} 1 & 3a \\ 2 & 2 \end{array} \right|}=\dfrac{1}{\left| \begin{array}{rr} 2 & 2 \\ -3 & -(a+3) \end{array} \right|}$$
This gives you,
$$\dfrac{\beta^2}{-8a+3}=\dfrac{\beta}{2-6a}=\dfrac{1}{-2a}$$
This yields, on eliminating $\beta$, $$\left(\dfrac{2-6a}{-2a}\right)^2=\dfrac{-8a+3}{-2a}$$ This simplifies to the following, $$10a^2-9a+2=0$$ whose roots are $\dfrac{2}{5}$ and $\dfrac{1}{2}$ which implies, the least $a$ is $\dfrac{2}{5}$. So, this completes your answer.