#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# Find the equation of the plane passing through the point (–1, 3, 2) and perpen- dicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0 ?

SHAIK AASIF AHAMED
6 years ago
Hello student,
Given equation of plane is passing through the point (-1,3,2)
Hence a(x+1)+b(y-3)+c(z-2)=0
Since the plane isperpen- dicular to each ofthe planes x+ 2y + 3z = 5 and 3x + 3y + z = 0
A+2B+3C=0 and 3A+3B+c=0
$\frac{A}{2-9}=\frac{B}{9-1}=\frac{C}{3-6}=k$
We get A=-7k,B=8k,c=-3k
So the required equation of plane is 7x-8y+3z+25=0
Thanks and Regards
Shaik Aasif