# The equation of the plane containing the line and parallel to the plane, Option 1) Option 2) Option 3) Option 4)

As we learnt in

Equation of any plane passing through the line of intersection of two planes (Cartesian form ) -

The equation of any plane passing through the line of intersection of two planes

$ax+by+cz+d= 0$ and

$a_{1}x+b_{1}y+c_{1}z+d_{1}= 0$ is given by

$\left ( ax+by+cz+d \right )+\lambda \left ( a_{1}x+b_{1}y+c_{1}z+d _{1}\right )= 0$

-

Equation of a plane be $\left ( 2x-5y+z-3 \right ) +\lambda \left ( x+y+4z-5 \right ) = 0$            ..........(i)

Since a plane is parallel to  x+3y+6z-1

$\frac{2+\lambda }{1} = \frac{\lambda - 5}{3}= \frac{1+4\lambda }{6}\Rightarrow \lambda = \frac{-11}{2}$

After substituting $\lambda = \frac{-11}{2} in (i)$

Hence requred plane will be

Option 1)

This option is incorrect

Option 2)

This option is incorrect

Option 3)

This option is correct

Option 4)

This option is incorrect

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