Solution:
The first observation is that we can find 5 pairs of squares such that the two numbers in a pair have the same parity. We can see this as follows:

Let us take such 5 pairs: say $\left(x_{1}^{2}, y_{1}^{2}\right),\left(x_{2}^{2}, y_{2}^{2}\right), \ldots,\left(x_{5}^{2}, y_{5}^{2}\right)$. Then $x_{j}^{2}-y_{j}^{2}$ is divisible by 4 for $1 \leq j \leq 5$. Let $r_{j}$ be the remainder when $x_{j}^{2}-y_{j}^{2}$ is divisible by $3, 1 \leq j \leq 5$. We have 5 remainders $r_{1}, r_{2}, r_{3}, r_{4}, r_{5}$. But these can be 0, 1 or 2. Hence either one of the remainders occurs 3 times or each of the remainders occurs once. If, for example, $r_{1}=r_{2}=r_{3}$, then 3 divides $r_{1}+r_{2}+r_{3}$; if $r_{1}=0, r_{2}=1$ and $r_{3}=2$, then again 3 divides $r_{1}+r_{2}+r_{3}$. Thus we can always find three remainders whose sum is divisible by 3. This means we can find 3 pairs, say, $\left(x_{1}^{2}, y_{1}^{2}\right),\left(x_{2}^{2}, y_{2}^{2}\right),\left(x_{3}^{2}, y_{3}^{2}\right)$ such that 3 divides $\left(x_{1}^{2}-y_{1}^{2}\right)+\left(x_{2}^{2}-y_{2}^{2}\right)+\left(x_{3}^{2}-y_{3}^{2}\right)$. Since each difference is divisible by 4, we conclude that we can find 6 numbers $a^{2}, b^{2}, c^{2}, d^{2}, e^{2}, f^{2}$ such that
$$a^{2}+b^{2}+c^{2} \equiv d^{2}+e^{2}+f^{2} \quad(\bmod 12)$$