 Solution

Call the three planes $P_1,P_2,P_3$ in their given order.

For $\alpha \ne 3,-1/2$, the nonzero determinant makes their normal vectors linearly independent. The three planes meet at the single point found in part (c)(i), for every $\beta$.

For $\alpha =3$, $P_2$ and $P_3$ coincide. Their common plane and $P_1$ have nonparallel normal vectors, since their first two coefficients are $(1,-1)$ and $(1,6)$. Thus they intersect in the straight line from part (c)(ii), for every $\beta$.

For $\alpha =-1/2$, write the three plane equations asThe last two meet on $z=0$, $x-y=-1$, which never lies on $P_1$. If $\beta =3$, $P_1$ and $P_2$ are distinct parallel planes, while $P_3$ cuts both. If $\beta =-1/2$, $P_1$ and $P_3$ are distinct parallel planes, while $P_2$ cuts both. For every other $\beta$, each pair of planes intersects in a straight line; these three pairwise intersection lines are distinct and parallel to $(1,1,0{)}^T$, with no common point. These exhaust the geometric cases.