If the equation of plane passing

Question:

If the equation of plane passing through the mirror image of a point $(2,3,1)$ with respect to line $\frac{x+1}{2}=\frac{y-3}{1}=\frac{z+2}{-1}$ and containing the line $\frac{x-2}{3}=\frac{1-y}{2}=\frac{z+1}{1}$ is $\alpha x+\beta y+\gamma z=24$, then $\alpha+\beta+\gamma$ is equal to :

  1. 20

  2. 19

  3. 18

  4. 21


Correct Option: , 2

Solution:

Line $\frac{x+1}{2}=\frac{y-3}{1}=\frac{z+2}{-1}$

$\overrightarrow{\mathrm{PM}}=(2 \lambda-3, \lambda,-\lambda-3)$

$\overrightarrow{\mathrm{PM}} \perp(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})$

$4 \lambda-6+\lambda+\lambda+3=0 \Rightarrow \lambda=\frac{1}{2}$

$\therefore \mathbf{M} \equiv\left(0, \frac{7}{2}, \frac{-5}{2}\right)$

$\therefore$ Reflection $(-2,4,-6)$

Plane : $\left|\begin{array}{ccc}x-2 & y-1 & z+1 \\ 3 & -2 & 1 \\ 4 & -3 & 5\end{array}\right|=0$

$\Rightarrow(x-2)(-10+3)-(y-1)(15-4)+(z+1)(-1)=0$

$\Rightarrow-7 x+14-11 y+11-z-1=0$

$\Rightarrow 7 x+11 y+z=24$

$\therefore \alpha=7, \beta=11, \gamma=1$

$\alpha+\beta+\gamma=19 \quad$ Option (2)

 

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