Cicumcircle in 3D ??

Let the three points be on the plane $ax + by + cz = d$ . Then:

$(1, 0, 0) \rightarrow a = d$

$(3, 1, 0) \rightarrow 3a + b = d$

$(5, 2, 1) \rightarrow 5a + 2b + c = d$

These three equations solve to $b = -2a$ , $c = 0$ , and $d = a$ , so the equation of the plane of the circle is $x - 2y = 1$ .

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Method 1:

Let $P$ be $(1, 0, 0)$ , $Q$ be $(3, 1, 0)$ , and $S$ be $(5, 2, 1)$ .

The midpoint between $P$ and $Q$ is $(2, \frac{1}{2}, 0)$ and $\overrightarrow{PQ} = (2, 1, 0)$ , so the equation of the plane perpendicular to $PQ$ through its midpoint is $4x + 2y = 9$ .

The midpoint between $P$ and $S$ is $(3, 1, \frac{1}{2})$ and $\overrightarrow{PR} = (4, 2, 1)$ , so the equation of the plane perpendicular to $PS$ through its midpoint is $8x + 4y + 2z = 29$ .

The intersection of these three planes is the center $(C_x, C_y, C_z)$ , which solves to $C_x = 2$ , $C_y = \frac{1}{2}$ , and $C_z = \frac{11}{2}$ .

The radius $R$ is the distance between the center and one of the points, say $P$ , so $R^2 = (C_x - P_x)^2 + (C_y - P_y)^2 + (C_z - P_z)^2 = \frac{63}{2}$ .

Therefore, $C_x + C_y + C_z + 2R^2 = \boxed{71}$ .

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Method 2:

The center $(C_x, C_y, C_z)$ would also be on the plane of the circle (see above), so:

$C_x - 2C_y = 1$

The three points are also a distance of $R$ away from the center, so:

$(1, 0, 0) \rightarrow (C_x - 1)^2 + C_y^2 + C_z^2 = R^2$

$(3, 1, 0) \rightarrow (C_x - 3)^2 + (C_y - 1)^2 + C_z^2 = R^2$

$(5, 2, 1) \rightarrow (C_x - 5)^2 + (C_y - 2)^2 + (C_z - 1)^2 = R^2$

These four equations solve to $C_x = 2$ , $C_y = \frac{1}{2}$ , $C_z = \frac{11}{2}$ , and $R^2 = \frac{63}{2}$ , so $C_x + C_y + C_z + 2R^2 = \boxed{71}$ .