A) \[x+3y+6z=7\]
B) \[x+6y+12z=-13\]
C) \[2x+6y+12z=13\]
D) \[x+3y+6z=-7\]
Correct Answer: A
Solution :
Equation of the plane containing the line \[2x-5y+7-3=0=x+y+47-5\] is\[(2x-5y+7-3)+\lambda (x+y+47-7)=0\] \[\Rightarrow \]\[(2+\lambda )x+(\lambda -5)y+(4\lambda +1)7-(3+5\lambda )=0\]_______(1) As equation (1) is parallel to the plane \[x+3y+6z=1\] then \[\frac{2+\lambda }{1}=\frac{\lambda -5}{3}=\frac{4\lambda +1}{6}\] ?(2) from (2) \[\lambda =\frac{-11}{2}\] using \[\lambda =\frac{-11}{2}\] in equation (1) we have equation of plane is \[x+3y+6z=7\]
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