d
By equation $1$ and $3$
And $\begin{array}{c}y+2 z=8 \\ y=8-2 z \\ x=-2+z\end{array}$
Now putting in equation $2$
$\alpha(z-2)+\beta(-2 z+8)+7 z=3$
$\Rightarrow(\alpha-2 \beta+7) z=2 \alpha-8 \beta+3$
So equations have unique solution if $\alpha-2 \beta+7 \neq 0$
And equations have no solution if $\alpha-2 \beta+7=0$ and $2 \alpha-8 \beta+3 \neq 0$
And equations have infinite solution if $\alpha-2 \beta+7=0$ and $2 \alpha-8 \beta+3=0$