Find the distance of the point `(2, 3, 4)` from the plane `3x + 2y + 2z + 5= 0` measured parallel to the line `(x+3)/3=(y-2)/6=z/2` .

Explanation

Let l be the given below
`l : (x+3)/(3) = (y-2)/(6) = z/2` and the given point be ` P(2,3,4)` . Let the given point be `P(2,3,4)` Let a line parallel to line 'l' through P meets the plane at Q.
Equation of line PQ
`(x-2)/(3) = (y-3)/(6) = (z-4)/(2) = lambda` (say)
Co-ordinates of point Q lies on PQ are
`(3lambda+2,6lambda+3,2lambda+4)`
Point Q lies on the plane `3x+2y+2z+5=0`
`:. 3(3lambda+2)+2(6lambda+3)+2(2lambda+4)+5=0`
`rArr 9lambda+6+12lambda +6+4lambda+8+5=0`
`rArr 45lambda+25=0`
`rArr lambda = -1`
`:.` Co-ordinates of `Q = (-1,-3,2)`
and `PQ = sqrt((2+1)^(2)+(3+3)^(2)+(4-2)^(2))`
`= sqrt(9+36+4) = 7` units.