The vector equation of the plane through the line of intersection

Question:

The vector equation of the plane through the line of intersection of the planes $x+y+z=1$ and $2 x+3 y+4 z=5$ which is perpendicular to the plane $x-y+z=0$ is :

  1. $\overrightarrow{\mathrm{r}} \times(\hat{\mathrm{i}}+\hat{\mathrm{k}})+2=0$

  2. $\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}-\hat{\mathrm{k}})-2=0$

  3. $\overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}-\hat{\mathrm{k}})+2=0$

  4. $\overrightarrow{\mathrm{r}} \times(\hat{\mathrm{i}}-\hat{\mathrm{k}})+2=0$


Correct Option: , 3

Solution:

Let the plane be

$(x+y+z-1)+\lambda(2 x+3 y+4 z-5)=0$

$\Rightarrow(2 \lambda+1) \mathrm{x}+(3 \lambda+1) \mathrm{y}+(4 \lambda+1) \mathrm{z}-(5 \lambda+1)=0$

$\perp$ to the plane $x-y+z=0$

$\Rightarrow \lambda=-\frac{1}{3}$

$\Rightarrow$ the required plane is $\mathrm{x}-\mathrm{z}+2=0$

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