Question

The equation of the plane containing the lines 2x - 5y + z = 3, x + 4z = 5 and parallel to the plane x + 3y + 6z = 1, is

A
2x+6y+12z=13
B
x+3y+6z=7
C
x+3y+6z=7
D
2x+6y+12z=13

Solution

The correct option is D $$x + 3y + 6z = 7$$Let equation of plane containing the lines $$2x - 5y + z = 3$$ and $$x+y+4z=5$$ be $$\left( {2x - 5y + z - 3} \right) + \lambda \left( {x + y + 4z - 5} \right) = 0$$or, $$\left( {2 + \lambda } \right)x + \left( {\lambda - 5} \right)y + \left( {4\lambda + 1} \right)z - 3 - 5\lambda = 0........\left( i \right)$$This planes parallel to the plane $$x+3y+6z=1$$$$\therefore \frac{{2 + \lambda }}{1} = \frac{{\lambda - 5}}{3} = \frac{{4\lambda + 1}}{6}$$Considering first two qualities, we get$$\begin{array}{l} \Rightarrow 6+3\lambda =\lambda -5 \\ \Rightarrow 2\lambda =-11 \\ \Rightarrow \lambda =\frac { { -11 } }{ 2 } \end{array}$$Considering the last two qualities ,we get$$\begin{array}{l} \Rightarrow 6\lambda -30=3+12\lambda \\ \Rightarrow -6\lambda =33 \\ \Rightarrow \lambda =\frac { { -11 } }{ 2 } \end{array}$$So, the equation of required plane is $$\left( {2 - \frac{{11}}{2}} \right)x + \left( {\frac{{ - 11}}{2} - 5} \right)y + \left( {\frac{{ - 44}}{2} + 1} \right)z - 3 + 5 \times \frac{{11}}{2} = 0$$$$\begin{array}{l} \Rightarrow \frac { { -7 } }{ 2 } x-\frac { { 21 } }{ 2 } y-\frac { { 42 } }{ 2 } z+\frac { { 49 } }{ 2 } =0 \\ \Rightarrow x+3y+6z-7=0 \\ \Rightarrow x+3y+6z=7 \end{array}$$Hence,Option $$C$$ is correct answerMaths

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