Any point on x2+y2=1,z=0 is p(cosθ,sinθ,0) If foot of perpendicular of p on the plane 2x+3y+z=6 is (h,k,l) then
h−cosθ
2
=
k−sinθ
3
=
l−0
1
=−(
2cosθ+3sinθ+0−6
22+32+12
)=r( let ) h=2r+cosθ,k=3r+sinθ,l=r Hence, h−2l=cosθ and k−3l=sinθ Hence, (h−2l)2+(k−3l)2=1 When l=6−2h−3k Hence required locus is (x−2(6−2x−3y))2+(y−3(6−2x−3y))2=1 ⇒(5x+6y−12)2+4(3x+5y−9)2=1,z=6−2x−3y