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(C) The length of the perpendicular $AN$ from $A(4, 3, 1)$ to the plane $x - y + 2z + 3 = 0$ is given by $AN = \frac{|4 - 3 + 2(1) + 3|}{\sqrt{1^2 + (

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$7$

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(C) The length of the perpendicular $AN$ from $A(4, 3, 1)$ to the plane $x - y + 2z + 3 = 0$ is given by $AN = \frac{|4 - 3 + 2(1) + 3|}{\sqrt{1^2 + (-1)^2 + 2^2}} = \frac{6}{\sqrt{6}} = \sqrt{6}$.The coordinates of $N$ are found using $\frac{x-4}{1} = \frac{y-3}{-1} = \frac{z-1}{2} = -\frac{4-3+2+3}{6} = -1$. Thus,$x=3, y=4, z=-1$,so $N(3, 4, -1)$.Since $B(5, \alpha, \beta)$ lies on the plane $x - y + 2z + 3 = 0$,we have $5 - \alpha + 2\beta + 3 = 0$,which implies $\alpha = 2\beta + 8$.The area of $\Delta ABN = \frac{1}{2} 	imes AN 	imes BN = 3\sqrt{2}$. Substituting $AN = \sqrt{6}$,we get $\frac{1}{2} 	imes \sqrt{6} 	imes BN = 3\sqrt{2}$,so $BN = \frac{6\sqrt{2}}{\sqrt{6}} = 2\sqrt{3}$.$BN^2 = (5-3)^2 + (\alpha-4)^2 + (\beta+1)^2 = 4 + (2\beta+8-4)^2 + (\beta+1)^2 = 4 + (2\beta+4)^2 + (\beta+1)^2 = 12$.$4 + 4\beta^2 + 16\beta + 16 + \beta^2 + 2\beta + 1 = 12 \implies 5\beta^2 + 18\beta + 9 = 0$.Factoring gives $(5\beta + 3)(\beta + 3) = 0$. Since $\beta \in \mathbb{Z}$,we have $\beta = -3$. Then $\alpha = 2(-3) + 8 = 2$.Finally,$\alpha^2 + \beta^2 + \alpha\beta = (2)^2 + (-3)^2 + (2)(-3) = 4 + 9 - 6 = 7$.

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