Square of Distance from Point to Plane of Consistency

To find the square of the distance $D$ from the point $(0, 1, 0)$ to the plane $P$, we first determine the condition for the system of linear equations to be consistent.

Since the rows of the coefficient matrix satisfy $R_1 + R_3 = 2R_2$, consistency requires the same linear relation to hold for the constants: $\alpha + (\gamma - 1) = 2\beta \implies \alpha - 2\beta + \gamma = 1$

Thus, the plane $P$ containing all points $(\alpha, \beta, \gamma)$ is given by the equation: $x - 2y + z - 1 = 0$

The perpendicular distance $d$ from the point $(0, 1, 0)$ to the plane $x - 2y + z - 1 = 0$ is: $d = \frac{|1(0) - 2(1) + 1(0) - 1|}{\sqrt{1^2 + (-2)^2 + 1^2}} = \frac{3}{\sqrt{6}}$

The square of the distance $D = d^2$ is therefore: $D = \left(\frac{3}{\sqrt{6}}\right)^2 = \frac{9}{6} = \frac{3}{2} = 1.5$