Reduction of a quadratic form with cross terms

Approach 1: direct grouping. Use the identity
$$xy+yz+zx=\frac12\left[(x+y+z)^2-(x^2+y^2+z^2)\right].$$
This is not directly a sum of squares, but shows the signature.

Approach 2: substitution. Let $u=x+y$, $v=x-y$, $z=z$ (this is invertible: $x=(u+v)/2$, $y=(u-v)/2$). Then
$$xy=\frac{u^2-v^2}{4},\qquad yz=\frac{u-v}{2}z,\qquad zx=\frac{u+v}{2}z.$$
So
$$Q=\frac{u^2-v^2}{4}+uz.$$
Complete the square in $u$:
$$Q=\frac14(u+2z)^2-\frac14v^2-z^2.$$

Signature. One positive square, two negative squares. Signature $(1,2,0)$.

Conclusion. $Q=\frac14(u+2z)^2-\frac14v^2-z^2$, with signature $(1,2,0)$.