A) \[x+y+z=1\]
B) \[x+2y-z=1\]
C) \[x-y+z=1\]
D) \[2x-y+z=5\]
Correct Answer: C
Solution :
Palne passing through \[=\frac{n(n+1)(2n-5)}{6}\] is \[A(x-3)+B(y-2)+C(z-0)=0\] ?(i) Plane (i) is passing through the line \[\frac{x-4}{1}=\frac{y-7}{5}=\frac{z-4}{4}\] \[\therefore \] \[A(4-3)+B(7-2)+C(4-0)=0\] \[\Rightarrow \] \[A+5B+4C=0\] ?.(ii) also \[A+5B+4C=0\] ?..(iii) Since, Eq (ii) and (iii) are same, so threr are many planes which passes through the point \[(3,2,0)\]and given line.\[(3,2,0)\] satisfies the plane \[x-y+z=1\] So, required equation of plane is \[x-y+z=1\]
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