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Question

Find the equation of the plane passing through the intersection of the planes 3x+2yz+1=0 and x+y+z2=0 and the point (2,2,1) .

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Solution

Let us take z=0, then

xy=1

3x=3

x=1,y=1

Thus the point is (1,1,0)

when x=0, then

2yz=1

yz=2

Multiplying by 3, we get

3y3z=6

y+3z=1

2y=7

y=72,z=32

Thus the point is (0,72,32)

Let a be the vector from (1,2,1) to(1,1,0)

a=1(1),12,01=2,1,1

Let b be the vector from (1,2,1) to (0,7/2,3/2):

b=0(1),722,321=1,32,12

The normal to the plane is:

n=a×b=∣ ∣ ∣ ∣ijk21113212∣ ∣ ∣ ∣=1i2j+4k

Hence, the equation of plane is given\quad by

nxx0,yy0,zz0=0

1,2,4x(1),y2,z1=0

1(1+x)2(y2)+4(1)=0

x2y+4=1 which is the required equation of plane.

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