Integer solutions to $a^{2}+b^{2}+c^{2} = d^{2}+e^{2}+f^{2}$, where $a, b, c, d, e, f \neq 0$ How does one approach something like this? Is there an equivalent Legendre's three-square theorem for the sum of three squares in two different ways? 

It seems like the only way to approach it would be via some computer power.
There is a method to find the different primitive representations of an integer $n$ as the sum of three squares, see here. This gives solutions to the above equation. For example, we have $150$ different representations of $n=225=a^2+b^2+c^2$, and $96$ primitive ones among them. So we can choose each pair of these solutions to obtain
$$
a^2+b^2+c^2=n=d^2+e^2+f^2.
$$
When $ a = -d, b=-e, c=-f $ then it satisfies the equality.

Consider 

$$ (-99)^2+(-100)^2+(-101)^2 = (99)^2+(100)^2+(101)^2 $$

Similarly

$$ (-d)^2+(-e)^2+(-f)^2= d^2+e^2+f^2 $$

Where $ a≠b≠c≠d≠e≠f$

So infinite integral solutions are possible. (Not positive integral).