Let π be the plane passing through the point (3,−3,1) and perpendicular to the line joining the points (3,4,−1) and (2,−1,5). If the equation of the plane containing the points (3,4,−1),(−1,2,5) and perpendicular to the plane π is ax+y+cz−d=0 then 3(a+c)=
Plane π passes through (3,−3,1) and perpendicular to the line joining the points (3,4,−1) and (2,−1,5). ∴ DR's of normal to the plane are (3−2),(4,+1),‌(−1−5)≡(1,5,−6)
Equation of plane π is given by x+5y−6z+d=0 π consists the point (3,−3,1) ⇒3−15−6+d=0⇒d=18 ∴π≡x+5y−6z+18=0 Equation of plane containing points (3,4,−1) and (−1,2,5) is ax+y+cz−d=0...(i)
Normal to this plane will be perpendicular to the line joining the points (3,4,−1) and (−1,2,5). Then, a(3+1)+1(4−2)+c(−1−5)=0 ⇒4a+2−6c=0 ⇒2a+1−3c=0...(ii) Also, planes π and Eq.(i) are perpendicular. ⇒a+5−6c=0 ⇒a=6c−5...(iii) On putting a=6c−5 in Eq.(ii), 2(6c−5)+1−3c=0 ⇒12c−10+1−3c=0 ⇒9c=9 ⇒c=1 On putting c=1 in Eq.(iii), a=6−5=1 Thus, a=c=1 On putting a=c=1 in Eq.(i), x+y+z−d=0 This plane passes through (−1,2,5) ⇒−1+2+5−d=0 ⇒d=6 Now, 3(a+c) =3(1+1)=3×2 ∴3(a+c)=d