A ray of light passing through the point A(1, | vedclass

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(C) Let the image of point $A(1, 2, 3)$ across the plane $x+y+z-12=0$ be $S(p, q, r)$.Using the formula for the image of a point $(x_1, y_1, z_1)$ in

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$\sqrt{410}$

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(C) Let the image of point $A(1, 2, 3)$ across the plane $x+y+z-12=0$ be $S(p, q, r)$.Using the formula for the image of a point $(x_1, y_1, z_1)$ in the plane $ax+by+cz+d=0$:$\frac{p-1}{1} = \frac{q-2}{1} = \frac{r-3}{1} = \frac{-2(1+2+3-12)}{1^2+1^2+1^2} = \frac{-2(-6)}{3} = 4$.Thus,$p-1=4 \Rightarrow p=5$,$q-2=4 \Rightarrow q=6$,$r-3=4 \Rightarrow r=7$.So,the image point is $S(5, 6, 7)$.The reflected ray passes through $C(3, 5, 9)$ and appears to originate from $S(5, 6, 7)$. The line $SC$ is given by:$\frac{x-5}{3-5} = \frac{y-6}{5-6} = \frac{z-7}{9-7} \Rightarrow \frac{x-5}{-2} = \frac{y-6}{-1} = \frac{z-7}{2} = \lambda$.So,$x = 5-2\lambda$,$y = 6-\lambda$,$z = 7+2\lambda$.Since $B$ lies on the plane $x+y+z=12$:$(5-2\lambda) + (6-\lambda) + (7+2\lambda) = 12 \Rightarrow 18 - \lambda = 12 \Rightarrow \lambda = 6$.Substituting $\lambda=6$ into the coordinates of $B$:$x = 5-12 = -7$,$y = 6-6 = 0$,$z = 7+12 = 19$.Thus,$B = (-7, 0, 19)$.The distance $OB$ from the origin $O(0, 0, 0)$ is:$OB = \sqrt{(-7)^2 + 0^2 + 19^2} = \sqrt{49 + 361} = \sqrt{410}$.

$A$ ray of light passing through the point $A(1, 2, 3)$ strikes the plane $x+y+z=12$ at $B$ and on reflection passes through $C(3, 5, 9)$,then $OB$ is equal to

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