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 $(3)$.
 A: I think not to introduce additional equations, and directly solve the system of equations.
$$\left\{\begin{aligned}&R^2+Q^2+T^2=X^2+Y^2+Z^2\\&R^4+Q^4+T^4=X^4+Y^4+Z^4\end{aligned}\right.$$
Using integer parameters $k,s,t$ - Will make a replacement.
$$a=3(k+s-t)^2+k(k-t)$$
$$b=3(k+s-t)^2+s(s-t)$$
$$c=3(k+s-t)^2-t^2+(k+s)t-2ks$$
$$x=3(k+s-t)^2-ks$$
$$y=3(k+s-t)^2-t^2+(k+s)t-ks$$
$$z=3(k+s-t)^2+k^2+s^2-(k+s)t$$
Then the solution can be written as:
$$R=3a^4+(4a-b)b^3+(4a-c)c^3-(4a-x)x^3-(4a-y)y^3-(4a-z)z^3$$
$$Q=(4b-a)a^3+3b^4+(4b-c)c^3-(4b-x)x^3-(4b-y)y^3-(4b-z)z^3$$
$$T=(4c-a)a^3+(4c-b)b^3+3c^4-(4c-x)x^3-(4c-y)y^3-(4c-z)z^3$$
$$X=(4x-a)a^3+(4x-b)b^3+(4x-c)c^3-3x^4-(4x-y)y^3-(4x-z)z^3$$
$$Y=(4y-a)a^3+(4y-b)b^3+(4y-c)c^3-(4y-x)x^3-3y^4-(4y-z)z^3$$
$$Z=(4z-a)a^3+(4z-b)b^3+(4z-c)c^3-(4z-x)x^3-(4z-y)y^3-3z^4$$
To obtain relatively Prime solutions - after substitution should be reduced to common divisor.
A: (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.
