JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Vector Algebra

Since the four points are coplanar, the vectors $\vec{b}-\vec{a}$, $\vec{c}-\vec{a}$ and $\vec{d}-\vec{a}$ lie in one plane, so their scalar triple product vanishes: $[\vec{b}-\vec{a}\vec{c}-\vec{a}\vec{d}-\vec{a}]=0$. Expanding and cancelling every box that repeats a vector, $[\vec{b}\vec{c}\vec{d}]-[\vec{b}\vec{c}\vec{a}]-[\vec{b}\vec{a}\vec{d}]-[\vec{a}\vec{c}\vec{d}]=0$. Rewriting each remaining box by cyclic rearrangement gives $[\vec{a}\vec{b}\vec{c}]=[\vec{d}\vec{c}\vec{a}]+[\vec{b}\vec{d}\vec{a}]+[\vec{c}\vec{d}\vec{b}]$.