Let A,B,C be points with position vectors `2hati-hatj+hatk,hati+2hatj+3hatkand 3hati+hatj+2hatk` respectively. Find the shortest distance between point B and plane OAC.

Explanation

Shortest distance between B and plane OAC is perpendicular distance of point B from the plane BM, M is foot of perpendicualar from B on plane OAC.
Now BM= projection vector `vec(AB)` on vector perpendicualar to the plane. Now vector perpendicular to the plane is `vec(OA)xx vec(OC) = veca xx vecc`. Therefore,
`BM=(vec(OB).(vecaxxvecc))/(|vecaxxvecc|)`
`=(vecb.(vecaxxvecc))/(|vecaxxvecc|)`
`vecaxxvecc=|{:(hati,hatj,hatk),(2,-1,1),(3,1,2):}|=-3hati-hatj+5hatk`
`vecb.(vecaxxvecc)=(hati+2hatj+3hatk).(-3hati-hatj+5hatk)=-3-2+15=10`
`|vecaxxvecc|=|-3hati-hatj+5hatk|=sqrt(9+1+25)=sqrt35`
`BM = 10/sqrt35=2sqrt(5/7)`