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Question

What is the equation of the plane passing through the line 3x−y−5z=2=x−2y+3z and perpendicular to the plane x−y+z=3?

A
2x+3y+z=2
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B
3x+2yz=2
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C
7(xz)=6
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D
19x8y27z=14
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Solution

The correct option is C 19x8y27z=14

Equation of the plane passing through the line 3xy5z=2=x2y+3z is

3xy5z2+k(x2y+3z2)=0

(3+k)x(1+2k)y+(5+3k)z=2+2k

Since, it is perpendicular to xy+z=3

Therefore, ((3+k)i(1+2k)j+(5+3k)k).(ij+k)=0

3+k+1+2k5+3k=0 k=16

Therefore, desired plane is (3+16)x(1+2(16))y+(5+3(16))z=2+2(16)

19x8y27=14


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