Step 1: Coplanarity condition.

$[\vec{X}\vec{Y}\vec{Z}]=0$ can be written as the product of a scalar coefficient determinant and the box product $[\vec{x}\vec{y}\vec{z}]$. Since $[\vec{x}\vec{y}\vec{z}]\ne 0$, we need

$$\left|\begin{array}{c}\alpha \beta -1\\ -1\alpha \beta \\ \beta -1\alpha \end{array}\right|=0.$$

Step 2: Expand.

${\alpha }^3+{\beta }^3+(-1{)}^3-3(\alpha )(\beta )(-1)=0\Rightarrow {\alpha }^3+{\beta }^3-1+3\alpha \beta =0$.

Step 3: Use the factorisation identity.

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ with $a=\alpha$, $b=\beta$, $c=-1$.

Since $\alpha ,\beta >0$ and $\alpha \ne \beta$, the quadratic factor is positive, so $\alpha +\beta -1=0$.

Step 4: Compute.

$\alpha +\beta -3=1-3=-2$.