Given three real numbers $x$, $y$, and $z$ such that $x^2+y^2+z^2=250$ and $xy+yz-zx=65$, determine the value of ${(x-y+z)}^2$.

The correct answer is 120.

To find the value of , we can use the algebraic expansion for the square of a trinomial:

${(a+b+c)}^2=a^2+b^2+c^2+2ab+2bc+2ca$

Substituting $a=x$, $b=-y$, and $c=z$ into the expansion formula:

${(x-y+z)}^2=x^2+{(-y)}^2+z^2+2\left(x\right)(-y)+2(-y)\left(z\right)+2\left(z\right)\left(x\right)$

Simplifying each term gives:

By factoring out $-2$ from the last three terms, we can regroup the expression as:

${(x-y+z)}^2=(x^2+y^2+z^2)-2(xy+yz-zx)$

We are given the values of the expressions in the parentheses:

$x^2+y^2+z^2=250$

$xy+yz-zx=65$

Substituting these values into our simplified expression:

${(x-y+z)}^2=250-2\left(65\right)$

${(x-y+z)}^2=250-130=120$

Thus, the value of ${(x-y+z)}^2$ is 120.