Find the equation of the plane through the pont A(1,2,1) and perpendicular to the line joining the points P(1,4,2) and Q(2,3,5). Also find the distance of this plane from the line `(x+3)/2=(y-5)/-1=(z-7)/-1`

Explanation

The general equation of a plane passing through the point A(1,2,1) is given by
`a(x-1)+b(y-2)+c(z-1)=0`
D.r 's of normal to the plane (i) are a,b,c
Dr s' of line PQ are (2-1) , (3-4) ,(5-2) . i.e 1,-1,3
Since the required plane is perpendicualr to line PQ so hte normal to htis plane must be parallel to PQ
`therefore (a)/(1) =(b)/(-1) =(c )/(3) =lambda(say) rarr a=lambda , b= lambda` and `c=3 lambda`
The give line is `(x+3)/(2)=(y-5)/(-1)=(z-7)/(-1)`
This line passes through th epoint (-3,5,7)
Distance between the plane (ii) and the line (iii)
=Distance of any point on this line form the plane
= Length of perpendicualr form (-3,5,7) on plane (ii)
`=|(-3-5+21-2)|/sqrt(1^(2)+(-1)^(2)+3^(2))=(11)/sqrt(11)=sqrt(11)` units
Hence the distance between the disired plane and the given line is `sqrt(11)` units