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(C) Let the vertices be $A(1,2,3), B(3,-1,5),$ and $C(4,0,-3)$.First, we calculate the direction ratios of the sides:$\overrightarrow{AB} = (3-1, -1-2

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$4|\gamma|$

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(C) Let the vertices be $A(1,2,3), B(3,-1,5),$ and $C(4,0,-3)$.First, we calculate the direction ratios of the sides:$\overrightarrow{AB} = (3-1, -1-2, 5-3) = (2, -3, 2)$$\overrightarrow{BC} = (4-3, 0+1, -3-5) = (1, 1, -8)$$\overrightarrow{AC} = (4-1, 0-2, -3-3) = (3, -2, -6)$Now, check for orthogonality:$\overrightarrow{AB} \cdot \overrightarrow{AC} = (2)(3) + (-3)(-2) + (2)(-6) = 6 + 6 - 12 = 0$.Since the dot product is $0$, $\overrightarrow{AB} \perp \overrightarrow{AC}$, which means $\angle A = 90^{\circ}$.In a right-angled triangle, the circumcenter is the midpoint of the hypotenuse $BC$.Circumcenter $(\alpha, \beta, \gamma) = \left(\frac{3+4}{2}, \frac{-1+0}{2}, \frac{5-3}{2}\right) = \left(\frac{7}{2}, -\frac{1}{2}, 1\right)$.Thus, $\alpha = \frac{7}{2}, \beta = -\frac{1}{2}, \gamma = 1$.Then, $|\alpha| + |\beta| = |\frac{7}{2}| + |-\frac{1}{2}| = \frac{7}{2} + \frac{1}{2} = 4$.Since $|\gamma| = |1| = 1$, we have $4 = 4|\gamma|$.

If the circumcenter of the triangle formed by the points $(1,2,3), (3,-1,5)$ and $(4,0,-3)$ is $(\alpha, \beta, \gamma)$, then $|\alpha|+|\beta|=$

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