JEE Mains Model Paper 15

Any plane through the intersection of the given planes is
x + 2y + 3z-4 + K (2x +y-z+ 5) = 0
i.e. 1l + 2K)x + (2 + K)y + (3-K)z + 5K - 4 = 0
This is ⊥ to 5x + 3y + 6z + 8 = 0
∴ 5 (1 + 2K) + 3 (2 + K) + 6 (3 - K) = 0
⇒ 29 + 7K = 0
∴ K = −

29

7

∴ required plane is
(1−

58

7

) x + (2−

29

7

) y + (3+

29

7

) z -

145

7

- 4 = 0
⇒ -51x - 15y + 50z - 173 = 0
⇒ 51x ++ 15y - 50z + 173 = 0