Show that one combination exists (in $^{11}C_6$, 1 to 11 integers) such that $a^2+b^2+c^2 = d^2+e^2+f^2$ is divisible by 12

Question

The question came in one of my tests, I could not solve it. Unfortunately, it wasn't a class test so there is no way I can ask for solution.

For completion, the full problem is:

Show that there exists at least one combination of 6 integers selected in 1 to 11 ($^{11}C_6$), such that $a^2+b^2+c^2 = d^2+e^2+f^2$ is divisible by $12$, where $a, b,c,d,e,f$ are the integers.

Answer 1

There must be some misundertanding of the question. In its current form, it is impossible.

Recall for any integer $n$,

$$n^2 \equiv \begin{cases} 0 \pmod 4, & n \text{ even }\\ 1 \pmod 4, & n \text{ odd } \end{cases} $$ In order for $a^2 + b^2 + c^2 = d^2 + e^2 + f^2$ to be divisible by $12$ and hence by $4$, all six numbers $a, b, c, d, e, f$ has to be even. However, there are only $5$ even numbers among $\{ 1, \ldots, 11 \}$.
The condition $a, b, c, d, e, f$ distinct is not compatible with the condition that sum of 3 squares is divisible by $12$.