Determine α and β so that the line x-y+z-3=0, 2x+α y+z+β=0…

Determine $\alpha$ and $\beta$ so that the line $x-y+z-3=0,\ 2x+\alpha y+z+\beta=0$ lies in the $xOy$ plane.

Step 1: Solution

The plane $z=0$ must belong to the pencil: $\lambda(x-y+z-3)+\mu(2x+\alpha y+z+\beta)\equiv z$. Comparing the $x$ and $z$ coefficients: $\lambda+2\mu=0$ and $\lambda+\mu=1$, so $\lambda=2$ and $\mu=-1$. Then the $y$ and constant terms require:

\[ -2-\alpha=0,\qquad -6-\beta=0 . \]

\[\alpha=-2,\qquad\beta=-6\]

Check: $2(x-y+z-3)-(2x-2y+z-6)=z$. The line is $x-y=3,\ z=0$.