<s>[INST] Sinha's Theorem for Equal Sums of Like Powers $x_1^7+x_2^7+x_3^7+\dots$ Sinha’s theorem can be stated as, excluding the trivial case $c = 0$, if,
$$(a+3c)^k + (b+3c)^k + (a+b-2c)^k = (c+d)^k + (c+e)^k + (-2c+d+e)^k\tag{1} $$
for $\color{blue}{\text{both}}$ $k = 2,4$ then,
$$a^k + b^k + (a+2c)^k + (b+2c)^k + (-c+d+e)^k = \\(a+b-c)^k + (a+b+c)^k + d^k + e^k + (3c)^k
\tag{2}$$
for $k = 1,3,5,7$.
The system $(1)$  be equivalently expressed as,
$$\begin{align}
x_1^k+x_2^k+x_3^k\, &= y_1^k+y_2^k+y_3^k,\quad \color{blue}{\text{both}}\; k = 2,4\\
x_1+x_2-x_3\, &= 2(y_1+y_2-y_3)\\
x_1+x_2-x_3\, &\ne 0\tag{3}
\end{align}$$
There are only two quadratic parameterizations known so far to $(3)$, namely,
$$(-5x+2y+z)^k + (-5x+2y-z)^k + (6x-4y)^k = \\(9x-y)^k + (-x+3y)^k + (16x-2y)^k\tag{4}$$
where $126x^2-5y^2 = z^2$ and,
$$(6x+3y)^k + (4x+9y)^k + (2x-12y)^k = \\(-x+3y+3z)^k + (-x+3y-3z)^k + (-6x-6y)^k\tag{5}$$
where $x^2+10y^2 = z^2$ found by Sinha and (yours truly). The square-free discriminants  are $D = 70, -10$, respectively.
Question: Any other solution for $(3)$ in terms of quadratic forms?
P.S. There are a whole bunch of elliptic curves that can solve $(3)$.
 [/INST] (Too long for a comment.)
I simplified your expression and found they are ternary quadratic forms. (Why didn't you just simplify them? Maple and Mathematica can do it easily.)  So,
$$R^n+Q^n+T^n = X^n+Y^n+Z^n,\quad for\; n =2,4\tag{1}$$
$$\begin{align}R =& -2 k^2 - 2 k s + s^2 + 3 k t - t^2\\
Q =&\;  k^2 - 2 k s - 2 s^2 + 3 s t - t^2\\
T =&\;  k^2 + 4 k s + s^2 - 3 k t - 3 s t + 2 t^2\\
X =&\;  k^2 + k s + s^2 - t^2\\
Y =&\; k^2 + k s + s^2 - 3 k t - 3 s t + 2 t^2\\
Z =& -2 k^2 - 2 k s - 2 s^2 + 3 k t + 3 s t - t^2
\end{align}$$
Unfortunately, this also satisfies,
$$R+Q+T =X+Y+Z = 0$$
or equivalently, since $(1)$ involves even powers,
$$R+Q-(-T) =X+Y-(-Z) = 0$$
a case prohibited by Sinha's theorem.
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