A) \[2x-y-2z-3=0\]
B) \[2x+y-2z-9=0\]
C) \[2x+y+2z-5=0\]
D) \[2x-y+2z+1=0\]
Correct Answer: D
Solution :
Where \[M\,({{x}_{1}},{{y}_{1}},{{z}_{1}})=(2,3,-1)\] Let, \[M(2,3,-1)\] is the foot of the perpendicular from the point P. The line PM is the normal to the plane. \[\therefore \] DR's of the normal are \[(4-2),\,(2-3),(1+1)\] \[=2,\,-1,\,2=a,b,c\] \[\therefore \] required palne, \[a(x-{{x}_{1}})+b(y-{{y}_{1}})+c(z-{{z}_{1}})=0\] \[2(x-2)+(-1)(y-3)+2(z+1)=0\] \[2x-y+2z+1=0\]You need to login to perform this action.
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