A) \[x+y+z=3\]
B) \[x+2y+3z=6\]
C) \[2x+3y+4z=9\]
D) \[3x+4y+5z=18\]
Correct Answer: B
Solution :
Equation of plane containing the line of intersection of planes is \[(x+y+z-6)+\lambda (2x+3y+4z-12)=0\] Since, it passes through the point \[(1,1,1)\]. \[\therefore \] \[(1+1+1-6)+\lambda (2+3+4-12)=0\] \[\Rightarrow \] \[-3+\lambda (-3)=0\] \[\Rightarrow \] \[\lambda =-1\] Hence, required equation of plane is \[(x+y+z-6)-(2x+3y+4z-12)=0\] ie, \[x+2y+3z=6\]You need to login to perform this action.
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