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The equation of the plane containing the lines \[2x-5y+z=3;x+y+4z=5\] and parallel to the plane \[x+3y+6z=1\].
\[\begin{align}
  & A)2x+6y+12z=13 \\
 & B)x+3y+6z=-7 \\
 & C)x+3y+6z=7 \\
 & D)2x+6y+12z=-13 \\
\end{align}\]

Answer Verified Verified
Hint: We know that the equation of the plane containing the lines \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0\] and \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0\] is equal to \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}+\lambda \left( {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}} \right)=0\]. So, we should write the equation of the plane containing the lines \[2x-5y+z=3;x+y+4z=5\]. We know that the planes \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0\] and \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0\] are said to be parallel, then \[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}\]. It is also given that the equation of the plane containing the lines \[2x-5y+z=3;x+y+4z=5\] and parallel to the plane \[x+3y+6z=1\]. By using the concept, we can find the value of \[\lambda \]. By using this value of \[\lambda \], we can find the equation of the plane containing the lines \[2x-5y+z=3;x+y+4z=5\] and parallel to the plane \[x+3y+6z=1\].

Complete step-by-step answer:
We know that the equation of the plane containing the lines \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0\] and \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0\] is equal to \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}+\lambda \left( {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}} \right)=0\].
Now we should find the plane containing the lines \[2x-5y+z=3;x+y+4z=5\]. \[\begin{align}
  & \Rightarrow 2x-5y+z+\lambda \left( x+y+4z \right)=3+5\lambda \\
 & \Rightarrow 2x-5y+z+\lambda x+\lambda y+4\lambda z=3+5\lambda \\
 & \Rightarrow \left( 2+\lambda \right)x+\left( \lambda -5 \right)y+\left( 4\lambda +1 \right)=3+5\lambda \\
 & \Rightarrow \left( 2+\lambda \right)x+\left( \lambda -5 \right)y+\left( 4\lambda +1 \right)-3-5\lambda =0....(1) \\
\end{align}\]
From the question, it is clear that the equation of the plane containing the lines \[2x-5y+z=3;x+y+4z=5\] and parallel to the plane \[x+3y+6z=1\].
We know that the planes \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0\] and \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0\] are said to be parallel, then \[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}\].
So, it is clear that the planes \[\left( 2+\lambda \right)x+\left( \lambda -5 \right)y+\left( 4\lambda +1 \right)-3-5\lambda =0\] and \[x+3y+6z=1\] are parallel.
\[\Rightarrow \dfrac{2+\lambda }{1}=\dfrac{\lambda -5}{3}=\dfrac{4\lambda +1}{6}.....(2)\]
So, let us assume
\[\Rightarrow \dfrac{2+\lambda }{1}=\dfrac{\lambda -5}{3}=\dfrac{4\lambda +1}{6}=k.....(3)\]
So, from equation (3), we can write
\[\Rightarrow \dfrac{2+\lambda }{1}=k\]
By using cross multiplication, we get
\[\Rightarrow 2+\lambda =k....(4)\]
Now, from equation (3), we get
\[\Rightarrow \dfrac{\lambda -5}{3}=k\]
By using cross multiplication, we get
\[\Rightarrow \lambda -5=3k....(5)\]
Now let us substitute equation (5) in equation (4), then we get
\[\begin{align}
  & \Rightarrow \lambda -5=3\left( 2+\lambda \right) \\
 & \Rightarrow \lambda -5=6+3\lambda \\
 & \Rightarrow 2\lambda =-11 \\
 & \Rightarrow \lambda =\dfrac{-11}{2}.....(6) \\
\end{align}\]
Now we will substitute equation (6) in equation (1), then we get
\[\begin{align}
  & \Rightarrow \left( 2-\dfrac{11}{2} \right)x+\left( -\dfrac{11}{2}-5 \right)y+\left( 4\left( \dfrac{-11}{2} \right)+1 \right)z-3-5\left( \dfrac{-11}{2} \right)=0 \\
 & \Rightarrow \left( \dfrac{4-11}{2} \right)x+\left( \dfrac{-11-10}{2} \right)y+\left( 2(-11)+1 \right)z-3+\dfrac{55}{2}=0 \\
 & \Rightarrow \left( \dfrac{-7}{2} \right)x-\dfrac{21y}{2}-21z+\dfrac{49}{2}=0 \\
 & \Rightarrow \dfrac{-7x-21y-42z+49}{2}=0 \\
 & \Rightarrow -7x-21y-42z+49=0 \\
 & \Rightarrow -x-3y-6z+7=0 \\
 & \Rightarrow x+3y+6z-7=0 \\
 & \Rightarrow x+3y+6z=7.....(7) \\
\end{align}\]
So, from equation (7) we will get that the equation of the plane containing the lines \[2x-5y+z=3;x+y+4z=5\] and parallel to the plane \[x+3y+6z=1\] is equal to \[x+3y+6z=7\].

So, the correct answer is “Option C”.

Note: Students have a misconception that that the planes \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0\] and \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0\] are said to be parallel, then \[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}=\dfrac{{{d}_{1}}}{{{d}_{2}}}\]. But we know that the planes \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0\] and \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0\] are said to be parallel, then \[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}\]. So, students should avoid this misconception.