The equation of a plane containing

Question:

The equation of a plane containing the line of intersection of the planes $2 x-y-4=0$ and $y+2 z-4=0$ and passing through the point $(1,1,0)$ is :

  1. $x-3 y-2 z=-2$

  2. $2 x-z=2$

  3. $x-y-z=0$

  4. $x+3 y+z=4$


Correct Option: , 3

Solution:

Let the equation of required plane be;

$(2 x-y-4)+\lambda(y+2 z-4)=0$

$\because$ This plane passes through the

point $(1,1,0)$ then $(2-1-4)+\lambda(1+0-4)=0$

$\Rightarrow \lambda=-1$

Then, equation of required plane is,

$(2 x-y-4)-(y+2 z-4)=0$

$\Rightarrow 2 x-2 y-2 z=0 \Rightarrow x-y-z=0$

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