= Solution
Call the three <planes> $P_1,P_2,P_3$ in their given order.
For $\alpha\ne3,-1/2$, 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 as
$$
P_1:x-y+\beta z=1,\qquad P_2:x-y+3z=-1,\qquad P_3:x-y-\tfrac12z=-1.
$$
The 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.
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