Find the distance of the point (2,3,4) from the plane `vecr.(3hati-6hatj+2hatk)+11=0`.

Explanation

We know that the perpendicular distance of a point with position vector `veca`, from the plane `vecr.vecn=q` is given by
`p=(|veca.vecn-q|)/(|vecn|)`.
Here, `veca=(2hati+3hatj+4hatk), vecn=(3hati-6hatj+2hatk)` and q=-11.
`therefore` the distance is given by
`p=(|(2hati+3hatj+4hatk).(3hati-6hatj+2hatk)-(11)|)/sqrt(3^(2)+(-6)^(2)+2^(2))`
`=(|(2hati+3hatj+4hatk).(3hati-6hatj+2hatk)+11|)/sqrt(49)`
`=(|(6-18+8+11)|)/(7)=7/7=1` unit.
Hence, the required distance is 1 unit.