Let the image of the point left(frac{5}{3}, f | vedclass

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(C) Let the point be $A = \left(\frac{5}{3}, \frac{5}{3}, \frac{8}{3}\right)$ and the plane be $x-2y+z-2=0$.The formula for the image $P(x, y, z)$ of

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$15$

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(C) Let the point be $A = \left(\frac{5}{3}, \frac{5}{3}, \frac{8}{3}\right)$ and the plane be $x-2y+z-2=0$.The formula for the image $P(x, y, z)$ of a point $(x_1, y_1, z_1)$ in the plane $ax+by+cz+d=0$ is given by $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = -2 \frac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}$.Substituting the values: $\frac{x-\frac{5}{3}}{1} = \frac{y-\frac{5}{3}}{-2} = \frac{z-\frac{8}{3}}{1} = -2 \frac{\frac{5}{3} - 2(\frac{5}{3}) + \frac{8}{3} - 2}{1^2+(-2)^2+1^2}$.Simplifying the numerator: $\frac{5}{3} - \frac{10}{3} + \frac{8}{3} - 2 = \frac{3}{3} - 2 = 1 - 2 = -1$.So,$\frac{x-\frac{5}{3}}{1} = \frac{y-\frac{5}{3}}{-2} = \frac{z-\frac{8}{3}}{1} = -2 \frac{-1}{6} = \frac{1}{3}$.Thus,$x = \frac{5}{3} + \frac{1}{3} = 2$,$y = \frac{5}{3} - \frac{2}{3} = 1$,$z = \frac{8}{3} + \frac{1}{3} = 3$.So,$P = (2, 1, 3)$.The distance between $P(2, 1, 3)$ and $Q(6, -2, \alpha)$ is $13$.Using the distance formula: $\sqrt{(6-2)^2 + (-2-1)^2 + (\alpha-3)^2} = 13$.$4^2 + (-3)^2 + (\alpha-3)^2 = 169$.$16 + 9 + (\alpha-3)^2 = 169$.$25 + (\alpha-3)^2 = 169$.$(\alpha-3)^2 = 144$.$\alpha-3 = \pm 12$.Since $\alpha > 0$,$\alpha-3 = 12 \Rightarrow \alpha = 15$ or $\alpha-3 = -12 \Rightarrow \alpha = -9$ (rejected).Therefore,$\alpha = 15$.

Let the image of the point $\left(\frac{5}{3}, \frac{5}{3}, \frac{8}{3}\right)$ in the plane $x-2y+z-2=0$ be $P$. If the distance of the point $Q(6, -2, \alpha)$,where $\alpha > 0$,from $P$ is $13$,then $\alpha$ is equal to $...........$.

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