Find $x+y+z$, where $x, y, z$ are edges of a parallelepiped

Question

A parallelepiped has its edges represented by $x$, $y$ and $z$, these are directly proportional to the numbers $3$, $4$ and $5$ respectively. It is also known that they are, in this order, in increasing arithmetic progression. The diagonal of this parallelepiped measures $10\sqrt{2}$. Then the sum $x+y+z$ is equal to?

Comments:

By hypothesis we have $x = 3m$, $y = 4n$ and $z = 5k$. I'm tried to write $(x,y,z) = (y - r, y , y+r)$ where $r$ is a reason of the arithmetic progression, then $x + y + z = 3y$. Moreover, $\sqrt{x^2+y^2+z^2} = 10\sqrt{2}$ and from this equality we have an equation in two of the variables. I am not able to fit the facts and find the value of one of the variables to find the others.

Answer 1

If $x=3m,$ $y=4m$ and $z=5m$ so $$9m^2+16m^2+25m^2=200,$$ which gives $$m=2.$$ Can you end it now?

Answer 2

$$×=3k,y=4k,z=5k$$ $$x^2+y^2+z^2=9k^2+16k^2+25k^2=200$$ $$50k^2=200, k^2=4, k=2, k\not=-2$$ $$×=6,y=8, z=10$$