The equation,

$$((2 p - 13 q + 11 u)^k+(11 p - 2 u)^k+(11 q + 13 u)^k =\\( 2 p - 13 q - 11 u)^k+( 11 p + 2 u)^k+( 11 q - 13 u)^k\tag1$$

for $k=1$ is,

$$x_1+12\,x_2+x_3 = y_1+12\,y_2+y_3$$

It is also true for $k=2,6$, if,

$$225 p^3 + 458 p^2 q + 587 p q^2 + 1392 q^3 = -3 (75 p + 464 q) u^2\tag2$$

An initial point is $p,q = -104,17.$ Hence $(2)$ can be easily turned into an elliptic curve, so there is an infinite number of integer solutions to $(1)$. In general, let,

$$\alpha = n+1\\ \beta = n-1$$

then,

$$(2 p - \alpha q +\beta u)^k + (\beta p - 2 u)^k+(\beta q + \alpha u)^k =\\ (2 p - \alpha q -\beta u)^k + (\beta p + 2 u)^k+(\beta q -\alpha u)^k\tag3$$

is true for $k=2,6$ if there is $p,q,n$ such that,

$$Poly_1:= (-3+n)(5-2n+n^2)p + 4n(1+n^2)q$$

$$Poly_2:= (-3+n)(5-2n+n^2)p^3 + 2(5+11n-5n^2+n^3)p^2q - (5+7n+15n^2-3n^3)pq^2 + 4n(1+n^2)q^3$$

and,

$$\color{red}{-}Poly_1 Poly_2 = \text{square}\tag4$$

A trivial solution is $q = \frac{(3-n)p}{2n}$ which yields,

$$\color{red}{-}Poly_1 Poly_2 = \frac{(-9+n^2)^2(-1+n)^4p^4}{4n^2}\tag5$$

P.S. Note that eqn $(3)$ also obeys,

$$x_1+nx_2+x_3 = y_1+ny_2+y_3$$

where the example was just the case $n=12$.

Questions:

1. For what other positive integer $n$ below a bound can we find solutions to non-zero $(4)$ or with $(3-n)p-2nq \neq 0$? (The constraint is to prevent trivial solutions. I have found $n=12, 15, 21, 30, 33, 135$ but I am not sure if this is exhaustive for $n<150$.)
2. Excepting $n=1$, is it true that all other $n$ are multiples of $3$?