The equation of the plane passing through the point (1,2,–3) and perpendicular to the planes 3x+y−2z=5 and 2x−5y−z=7 is:
A
6x−5y+2z+10=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
11x+y+17z+38=0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
6x−5y−2z−2=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
3x−10y−2z+11=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B11x+y+17z+38=0 ∵ Required plane is perpendicual to both the planes ∴Normal vector of required plane is →n=∣∣
∣
∣∣^i^j^k31−22−5−1∣∣
∣
∣∣ →n=−11^i−^j−17^k
Hence eqaution of required plane will be −11(x−1)−(y−2)−17(z+3)=0 ⇒11x+y+17z+38=0