Find the distance of the point P(6,5,9) from the plane determined by the points `A(3,-1,2),\ B(5,2,4)` and `C(-1,\ -1,\ 6)dot`

Explanation

Here
`(x_(1)=3, y_(1)=-1,z_(1)=2),(x_(2)=5,y_(2)=2,z_(2)=4)`.
and `(x_(3)=-1, y_(3)=-1, z_(3)=6)`.
The equation of the plane is ABC is given by
`|{:(x-x_(1),y-y_(1),z-z-(1)),(x_(2)-x_(1),y_(2)-y-(1),z_(2)-z_(1)),(x_(3)-x-(1),y_(3)-y_(1),z_(3)-z_(1)):}|=0`
`rArr |{:(x-3,y+1,z-2),(5-3,2+1,4-2),(-1-3,-1+1,6-2):}|=0 rArr |:{(x-3,y+1,z-2),(2,3,2),(-4,0,4):}|=0`
`rArr (x-3)(12-0)-(y+1)(8+8)+(z-2)(0+12)=0`
`rArr 12-16y+12z-76=0 rArr 3x-4y+3z-19=0`..........(i)
Distance of the point `P(6,5,9)` from the plane `3x-4y+3z-19=0`
=length of perpendicular from P(6,5,9) to the plane
`(3x-4y+3z-19=0`
`=(|(3 xx 6)-(4xx5)+(3xx9)-19)|)/(sqrt(3^(2)+(-4)^(2)+3^(2))=(|18-20+27-19|)/sqrt(34)`.
`6/sqrt(34) xx sqrt(34)/sqrt(34)=6/sqrt(34)=3/17 sqrt(34)` units.
Hence, the required distance is `3/17sqrt(34)` units.