A) \[2x+5y+z-8=0\]
B) \[x+y-z-1=0\]
C) \[2x+5y+z+4=0\]
D) \[x-y+z-1=0\]
Correct Answer: B
Solution :
Any plane passing through (1, 1, 1) is \[a(x-1)+b(y-1)+c(z-1)=0\] .....(i) Plane (i) is also passing through (1, ?1, ?1) \[\therefore a\,.\,0+b(-2)+c(-2)=0\] or, \[0.a-2b-2c=0\] or \[0.a-b-c=0\] or, \[0.a+b+c=0\] ......(ii) Plane (i) is perpendicular to \[2x-y+z+5=0\] So, \[2a-b+c=0\] .....(iii) From (ii) and (iii), \[a=b=1,\,c=-1\] Substituting in (i) we have \[x+y-z-1=0\].You need to login to perform this action.
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