How to find the number of possible planes equidistant from $5$ points $3$ of which are collinear.

Question

$\vec a,\vec b,\vec c,\vec d$ are four non zero vectors and points $P(\vec a), \; Q(\vec a+\vec b+\vec c), \; R(\vec a-\vec b-\vec c), \; S(\vec d),\; T(\vec a +\vec d)$ are distinct points such that T does not lie in the plane of P,Q and S. Find the number of planes equidistant from all the points $P,Q,R,S,T$.

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Any three vectors are coplaner, now $P,Q,R$ are collinear as, $\vec{OR}=\dfrac{2\vec{OP}-\vec{OQ}}{ 2-1}$. So any plane parallel to this line will be equidistant from $P,Q,R$. But I think that points $S,T$ may be aligned in space such that we get no plane which is equidistant from all five points. But answer is given $3$ planes are possible. Can not visualise at this stage. Please help.

Answer 1

$P,Q,R$ are collinear. Let $d=PQ$, with $R\in d $. Then there are $3$ cases to consider :

• Let $\mathcal P$ the plane containing $d$ and $S$, $S'$ the orthogonal projection of $S$ on $\mathcal P$, $I$ the midpoint of $SS'$ and $\mathcal P_1$ the plane parallel to $\mathcal P$ passing through $I $;
• Let $\mathcal Q$ the plane containing $d$ and $T$, $T'$ the orthogonal projection of $T$ on $\mathcal Q$, $J$ the midpoint of $TT'$ and $\mathcal P_2$ the plane parallel to $\mathcal Q$ passing through $J $;
• Let $e:=ST$ and $K$ the midpoint of the common perpendicular to $d$ and $e$ (see here for example) and let $\mathcal P_3$ the plane passing throug $K$, whose direction is the one spaned by that of $d$ and $e$.

$\mathcal P_1,\mathcal P_2,\mathcal P_3$ are the $3$ planes.