Q.Find the vector equation of the plane passing through the points A(1, -2, 1), B(2, -1, -3) and C(0, 1, 5).

Three points $A(\vec a)$, $B(\vec b)$, $C(\vec c)$ that do not lie on one line always determine exactly one plane. For a general point $P(\vec r)$ of that plane, the three chords $\overrightarrow{AP}$, $\overrightarrow{AB}$, $\overrightarrow{AC}$ all lie in the plane and are therefore coplanar, so their scalar triple product vanishes: $(\vec r-\vec a)\cdot\big[(\vec b-\vec a)\times(\vec c-\vec a)\big] = 0$. In effect, the two chords $\overrightarrow{AB}$ and $\overrightarrow{AC}$ supply the two in-plane directions that earlier concepts needed a separate $\vec b$ and $\vec c$ for, and their cross product $\overrightarrow{AB}\times\overrightarrow{AC}$ becomes the plane's normal. …