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?

Let $a=\frac{17}{15}$ and $b=\frac{4}{5}$. 
Hence, we get a value $\frac{2}{15}$.
Thus, it remains to prove that
$$(a-b)^2 + (2-a-b)^2 + (2a-3b)^2\geq\frac{2}{15}$$ or
$$10(3a-3b-1)^2+3(5b-4)^2\geq0$$
Done!
I got my solution by the following way.
We need to find a maximal $k$ for which the following inequality is true for all reals $a$, $b$ and $c$.
$$(a-b)^2 + (2-a-b)^2 + (2a-3b)^2\geq k$$ or
$$6a^2-4(3b+1)a+11b^2-4b+4-k\geq0,$$ for which we need
$$4(3b+1)^2-6(11b^2-4b+4-k)\leq0$$ or
$$15b^2-24b+10-3k\geq0,$$
for which we need $$12^2-15(10-3k)\leq0$$ or
$$k\leq\frac{2}{15}.$$
The equality occurs for $k=\frac{2}{15}$, $b=\frac{24}{2\cdot15}$, which is $b=\frac{4}{5}$ and for these values we obtain
$$(a-b)^2 + (2-a-b)^2 + (2a-3b)^2\geq \frac{2}{15}$$ it's
$$6a^2-4(3b+1)a+11b^2-4b+4-\frac{2}{15}\geq0$$ or
$$90a^2-60(3b+1)a+165b^2-60b+58\geq0$$ or
$$10(9a^2-6(3b+1)a+(3b+1)^2)-10(3b+1)^2+165b^2-60b+58\geq0$$ or
$$10(3a-3b-1)^2+75b^2-120b+48\geq0$$ or
$$10(3a-3b-1)^2+3(5a-4)^2\geq0.$$
