Find the distance of the point `P(3,8,2)` from the line `1/2(x-1)=1/4(y-3)=1/3(z-2)` measured parallel to the plane `3x+2y-2z+15=0.`

Options

A. 2
B. 3
C. 6
D. 7

Answer

D

Explanation

Let `P(3,8,2)` be the given point.
Let the equation of a line through `P(3,8,2)` and parallel to the plane `3x+2y-2z+15=0` be
`(x-3)/a=(y-8)/b=(z-2)/c` ………….i

Then `3a+2b-2c=0` .............ii
Line (i) intersects the line `(x-1)/2=(y-3)/4=(z-2)/3` at `Q`
`:.|(3-1,8-3,2-2),(a,b,c),(2,4,3)|=0implies15a-6b-2c=0`..............iii
Solving (ii) and (iii) by cross multiplication, we get
`a/(-16)=b/(-24)=c/(-48)` or `a/2=b/3=c/6`
Substituting a,b,c i (i) we get
`(x-3)/2=(y-8)/3=(z-2)/6` as the equation line `PQ`.
Let the coordinates of `Q` be `(2lamda+3,3lamda+8,6lamda+2)`.
It lies on the line `(x-1)/2=(y-3)/4=(z-2)/3`
`:.lamda+1=(3lamda+5)/4=2lamdaimplieslamda=1`
Hence `PQ=sqrt((5-3)^(2)+(11-8)^(2)+(8-2)^(2))=7`