L is a line passing through the point A(1,0,−3) and parallel to a line having direction ratios 0,1,−2.P is a point on the line L which is at a minimum distance from the plane 2x+3y+5z=1 Then, the equation of the plane through P and perpendicular to AP is
=z+3−2=λ Thus, any point on line L is of the form (1,λ,(−2λ−3)). The distance of (1,λ,(−2λ−3)) from the plane 2x+3y+5z=1 is, d=
|2(1)+3λ+5(−2λ−3)−1|
√22+32+52
=
|2+3λ−10λ−15−1|
√38
=
|7λ+14|
√38
Now, d will be minimum when 7λ+14=0 λ=−2 The coordinates of P are (1,-2,1) and DR 'S of AP are 0,-2,4 The equation of plane through P and perpendicular to AP is, −2y+4+4z−4=0 2y−4z=−8 y−2z+4=0