A(27, -243, 81) is a point in space. B, C, D | vedclass

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(D) The coordinates of point $A$ are $(x, y, z) = (27, -243, 81)$.The image of a point $(x, y, z)$ with respect to the $XY$-plane is $(x, y, -z)$. Thu

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-$45$

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(D) The coordinates of point $A$ are $(x, y, z) = (27, -243, 81)$.The image of a point $(x, y, z)$ with respect to the $XY$-plane is $(x, y, -z)$. Thus, $B = (27, -243, -81)$.The image of a point $(x, y, z)$ with respect to the $YZ$-plane is $(-x, y, z)$. Thus, $C = (-27, -243, 81)$.The image of a point $(x, y, z)$ with respect to the $ZX$-plane is $(x, -y, z)$. Thus, $D = (27, 243, 81)$.The centroid $(\alpha, \beta, \gamma)$ of triangle $BCD$ is given by the average of the coordinates of its vertices:$\alpha = \frac{27 - 27 + 27}{3} = \frac{27}{3} = 9$$\beta = \frac{-243 - 243 + 243}{3} = \frac{-243}{3} = -81$$\gamma = \frac{-81 + 81 + 81}{3} = \frac{81}{3} = 27$Therefore, $\alpha + \beta + \gamma = 9 - 81 + 27 = -45$.

$A(27, -243, 81)$ is a point in space. $B, C, D$ are images of $A$ with respect to $XY, YZ$, and $ZX$ planes respectively. If the centroid of the triangle $BCD$ is $(\alpha, \beta, \gamma)$, then $\alpha + \beta + \gamma =$

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