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 :
Correct Option: , 2
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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