Consider the planes P1:cy+bz=x P2:az+cx=y P3:bx+ay=z. P1,P2 and P3 pass through one line, if
A
a2+b2+c2=ab+bc+ca
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B
a2+b2+c2+2abc=1
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C
a2+b2+c2=1
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D
a2+b2+c2+2ab+2bc+2ca+2abc=1
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Solution
The correct option is Ba2+b2+c2+2abc=1 Given planes P1:cy+bz−x=0 P2:az+cx−y=0 P3:bx+ay−z=0
Above planes can be written as P1:−x+cy+bz=0...(1) P2:cx−y+az=0...(2) P3:bx+ay−z=0...(3)
This is a homogeneous system of equations.
As all planes pass through one line, it means this system has infinitely many solutions.
So, ∣∣
∣∣−1cbc−1aba−1∣∣
∣∣=0