Multiply $x^{2}+4 y^{2}+z^{2}+2 x y+x z-2 y z$ by $(-z+x-2 y)$.

(D) We have the expression: $(-z+x-2 y)(x^{2}+4 y^{2}+z^{2}+2 x y+x z-2 y z)$.
Rearranging the terms,we get: $(x-2 y-z)(x^{2}+(-2 y)^{2}+(-z)^{2}-(x)(-2 y)-(-2 y)(-z)-(x)(-z))$.
This expression is in the form $(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$,where $a=x$,$b=-2 y$,and $c=-z$.
Using the algebraic identity $(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca) = a^{3}+b^{3}+c^{3}-3abc$,we substitute the values:
$= (x)^{3} + (-2 y)^{3} + (-z)^{3} - 3(x)(-2 y)(-z)$.
$= x^{3} - 8 y^{3} - z^{3} - 6 x y z$.