Find the equation of the plane which passes through the point `(1,2,3)` and which is at the minimum distance from the point `(-1,0,2)dot`

Explanation

The plane passes through the point `A(1, 2, 3)` and is at the maximum distance from the point `B(-1, 0, 2)`, then the plane is perpendicular to line `AB`. Therefore, the direction ratios of the normal to the plane are 2, 2 and 1.
Hence, the equation of the plane is
`" "2(x-1)+2(y-2)+1(z-3)=0`
or `" "2x+2y+z=9`