Minimum of the given expression 
For all real numbers $a$ and $b$ find the minimum of the following expression.
  $$(a-b)^2 + (2-a-b)^2 + (2a-3b)^2$$

I tried expressing the entire expression in terms of a single function of $a$ and $b$. For example, if the entire expression reduces to $(a-2b)^2+(a-2b)+5$ then its minimum can be easily found. But nothing seems to get this expression in such a form, because of the third unsymmetric square.
Since there are two variables here we can also not use differentiation. 
Can you please provide hints on how to solve this?

\begin{eqnarray*}
(a-b)^2+(2-a-b)^2+(2a-3b)^2=6a^2-12ab+11b^2-4(a+b)+4 \\
=6\left(a-b-\frac{1}{3}\right)^2+5\left(b-\frac{4}{5}\right)^2+\color{red}{\frac{2}{15}}.
\end{eqnarray*}
