The real numbers $x,y$ and $z$ are such that $x-7y+8z=4$ and $8x+4y-z=7$. What is the maximum value $x^2-y^2+z^2?$ The real numbers $x,y$ and $z$ are such that 
$x-7y+8z=4$ and $8x+4y-z=7$. What is the maximum value of $x^2-y^2+z^2?$

From those equations I got:

$12z-5x=13y$

$12x+5z=13$

$12y+5=13z$

$12-5y=13x$

I know that $5,12,13$ is a pythag triplet but I don’t know what to do next. I think lagrange multipliers could be used but there should be a solution that doesn’t require calculus

Hints, suggestions and solutions would be appreciated.

Taken from the 2014 KIMC https://chiuchang.org/imc/wp-content/uploads/sites/2/2018/01/2014-IWYMIC-Individual.x17381.pdf
Solving the system
$$x-7y+8z=4$$
$$8x+4y-z=-7$$
we get
$$y=\frac{12}{5}-\frac{13}{5}x$$
$$z=\frac{13}{5}-\frac{12}{5}x$$
and we get
$$x^2-y^2+z^2=1$$
Very NICE!

Below please find an image of the surface $x^2-y^2+z^2=1$ together with the (thick red) line of intersection of the two given planes.
Solving for $x,y$ in terms of $z$

$$12x=13-5z$$

$$12y=13z-12$$

$$12^2(x^2-y^2+z^2)=(13-5z)^2-(13z-12)^2+144z^2$$

$$=-50z^2-(130+312)z+25$$

$$=-50\left(z+\dfrac{442}{100}\right)^2+25+50\left(\dfrac{442}{100}\right)^2$$

$$\le25+50\left(\dfrac{442}{100}\right)^2$$