Find the distance of the point `(1,-2,3)` from the plane `x-y+z=5` measured parallel to the line `(x-1)/2=(y-3)/3=(z+2)/-6`.

Explanation

The equation of the given plane is
`x-y+z=5`……………(i)
The equation of the given line is
`(x-1)/2=(y-3)/3=(z+2)/-6`………..(ii)
The equation of the line passing through the point `P(1,-2,3)` and parallel to the (ii) is given by
`(x-1)/(2)=(y+2)/3=(z-3)/-6=lambda` (say).
The general point on line (iii) is `(2lambda+1,3lambda-=2,6lambda+3)`.

For some value of `lambda`, let the point `Q(2lambda+1,3lambda-2,-6lambda+3)` lie on the plane (i), Then, we have
`(2lambda+1)-(3lambda-2)+(-6lambda+3)=5 rArr -7lambda=-1 rArr lambda=1/7`.
So, the coordinates of Q are
`Q(2/7+1,3/7-2,-6/7+3)`, i.e., `Q(9/7.-11/7,15/7)`.
Distance PQ `=sqrt((9/7-1)^(2)+(-11/7+2)^(2)+(15/7-3)^(2))`
`=sqrt((2/7)^(2)+(3/7)^(2)+(-6/7)^(2))=1/4sqrt(4+9+36)`
`=1/7 xx sqrt(49)=(1/7 xx 7)=1 `unit.
Hence, the requrired distance is 1 unit.