A) \[2x+3y+5z=1\]
B) \[2x-5y=4\]
C) \[5y-3z-3=0\]
D) \[3y+4z=0\]
Correct Answer: C
Solution :
Let the equation of plane containing the line \[\frac{x-2}{2}=\frac{y-3}{3}=\frac{z-4}{5}\]is \[a(x-2)+b(y-3)+c(z-4)=0\] Since, the normal to the plane is perpendicular to the above line: \[2a+3b+5c=0\] ... (i) Also, plane is parallel to x-axis \[a=0\] ... (ii) \[\therefore \]From Eqs. (i) and (ii), we get \[3b+5c=0\] \[\Rightarrow \] \[\frac{b}{5}=\frac{c}{-3}\] \[\therefore \] \[0(x-2)+5(y-3)-3(z-4)=0\] \[\Rightarrow \] \[5y-3z-3=0\]
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