consider a plane x+y-z=1 andpoint,4(1,2,-3).A line L has the equation x=1+3r,y=2-r and z=3+4rThe coordinate of a point B of line I such that AB is parallel to the plane isThe equation of the plane containing line L and point A has the equationThe distance between the points on the line which are at a distance of 4/3 from the plane is

The line x−13=y−2−1=z−34=rhence,AB→ is parallel to x+y−z=1⇒ 3r−r−4r−6=0 or r=−3B is (−8,5,−9)The equation of plane containing the line L is A(x−1)+B(y−2)+C(z−3)=0where  3A−B+4C=0----i(i) also contains point A(1,2,-3). Hence, C= 0and 3A=8.the equation of plane  x−1+3(y−2)=0  or x+3y−7=0  The distance of point (1 + 3r, 2-r,3 + 4r)from the plane is|1+3r+2−r−3−4r−1|1+1+1=|2r+1|3=43⇒ r=32,−52Hence, the points are A112,12,102 and B−132,92,−142⇒ AB=292