A) \[x-y+z=1\]
B) \[x+y+z=5\]
C) \[x+2y-z=1\]
D) \[2x-y+z=5\]
Correct Answer: A
Solution :
A line will be a plane, iff the normal to the plane is perpendicular to the line. a point on the line lies on the plane. Equation of any plane passing through (3 2 0) is \[a(x-3)+b(y-2)+c(z-0)=0\] Since, the line \[\frac{x-4}{1}=\frac{y-7}{5}=\frac{z-4}{4}\] lies on a plane. \[\therefore a(4-3)+b(7-2)+c(4-0)=0\] i.e., \[a+5b+4c=0\] .... (i) and \[a+5b+4c=0\] ... (ii) From Eqs. (i) and (ii), we get one of the solution is \[a=1,\,b=-1,c=1\] \[\therefore \] Required equation of plane is \[1\,(x-3)-1(y-2)+1(z-0)=0\] \[\Rightarrow \] \[x-y+z=1\]
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