A) \[x-2y-2z-3=0,\] \[2x+y-2z+3=0\]
B) \[x-2y+2z-3=0,\]\[2x+y+2z+3=0\]
C) \[x+2y-2z-3=0,\]\[2x-y-2z+3=0\]
D) None of these
Correct Answer: A
Solution :
Equation of planes passing through intersecting the planes \[3x-y-4z=0\] and\[x+3y+6=0\]is, \[(3x-y-4z)+\lambda (x+3y+6)=0\] \[(3+\lambda )+(3\lambda -1)-4z+6\lambda =0\] ......(i) Given, distances of plane (i) from origin is 1. \[\therefore \,\,\frac{6\lambda }{\sqrt{{{(3+\lambda )}^{2}}+{{(3\lambda -1)}^{2}}+{{(-4)}^{2}}}}=1\] or \[36{{\lambda }^{2}}=10{{\lambda }^{2}}+26\] or \[\lambda =\pm 1\] Put the value of \[\lambda \] in (i), \[\therefore \,(3x-y-4z)\pm (x+3y+6)=0\] or \[4x+2y-4z+6=0\] or \[2x+y-2z+3=0\] and \[2x-4y-4z-6=0\] or \[x-2y-2z-3=0\] Thus the required planes are \[x-2y-2z-3=0\]and\[2x+y-2z+3=0\].
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