A) \[x+3y+6z=7\]
B) \[2x+6y+12z=-13\]
C) \[2x+6y+12z=13\]
D) \[x+3y+6z=-7\]
Correct Answer: A
Solution :
[a] Equation of the plane containing the lines \[2x-5y+z=3\] and \[x+y+4z=5\] is \[2x-5y+z-3+\lambda (x+y+4z-5)=0\] \[\Rightarrow (2+\lambda )x+(-5+\lambda )y+(1+4\lambda )z+(-3-5\lambda )=0\] ?.(i) Since the plane (i) parallel to the given plane \[x+3y+6z=1\] \[\therefore \frac{2+\lambda }{1}=\frac{-5+\lambda }{3}=\frac{1+4\lambda }{6}\Rightarrow \lambda =-\frac{11}{2}\] Hence equation of the required plane is \[\left( 2-\frac{11}{2} \right)x+\left( -5-\frac{11}{2} \right)y+\left( 1-\frac{44}{2} \right)z+\left( -3+\frac{55}{2} \right)=0\]\[\Rightarrow (4-11)x+(-10-11)y+(2-44)z+(-6+55)=0\]\[\Rightarrow x+3y+6z=7\]
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