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(D) The normal to the line $Ax+By+C=0$ from the origin has the slope $m = \frac{B}{A}$.For the line $x+y+\sqrt{2}=0$,the normal line passes through $(

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$\frac{35 \pi}{12}$

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(D) The normal to the line $Ax+By+C=0$ from the origin has the slope $m = \frac{B}{A}$.For the line $x+y+\sqrt{2}=0$,the normal line passes through $(0,0)$ and is perpendicular to the given line. Its equation is $x-y=0$,so its slope is $1$. Thus,$	an \alpha = 1$. Since the normal is in the first or third quadrant,and we measure anti-clockwise from the positive $X$-axis,$\alpha = \frac{\pi}{4}$ or $\frac{5 \pi}{4}$. Given the normal vector $(1,1)$,$\alpha = \frac{\pi}{4}$.For the line $x-\sqrt{3}y-2=0$,the normal line is $\sqrt{3}x+y=0$,so its slope is $-\sqrt{3}$. Thus,$	an \beta = -\sqrt{3}$. The angle $\beta$ in the anti-clockwise direction is $\frac{2 \pi}{3}$.However,considering the normals as vectors from the origin,for $x+y+\sqrt{2}=0$,the normal vector is $(1,1)$,so $\alpha = \frac{\pi}{4}$.For $x-\sqrt{3}y-2=0$,the normal vector is $(1, -\sqrt{3})$,so $\beta = 2 \pi - \frac{\pi}{3} = \frac{5 \pi}{3}$.Sum $\alpha + \beta = \frac{\pi}{4} + \frac{5 \pi}{3} = \frac{3 \pi + 20 \pi}{12} = \frac{23 \pi}{12}$.Re-evaluating based on standard normal form $x \cos 	heta + y \sin 	heta = p$:Line $1: x+y+\sqrt{2}=0 \Rightarrow -\frac{1}{\sqrt{2}}x - \frac{1}{\sqrt{2}}y = 1$. Here $\cos \alpha = -\frac{1}{\sqrt{2}}, \sin \alpha = -\frac{1}{\sqrt{2}} \Rightarrow \alpha = \frac{5 \pi}{4}$.Line $2: x-\sqrt{3}y-2=0 \Rightarrow \frac{1}{2}x - \frac{\sqrt{3}}{2}y = 1$. Here $\cos \beta = \frac{1}{2}, \sin \beta = -\frac{\sqrt{3}}{2} \Rightarrow \beta = \frac{5 \pi}{3}$.Sum $\alpha + \beta = \frac{5 \pi}{4} + \frac{5 \pi}{3} = \frac{15 \pi + 20 \pi}{12} = \frac{35 \pi}{12}$.

If $\alpha$ and $\beta$ are the angles made by the normals drawn from the origin to the lines $x+y+\sqrt{2}=0$ and $x-\sqrt{3}y-2=0$ with the positive direction of the $X$-axis respectively,measured in the anti-clockwise direction,then $\alpha+\beta=$

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