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(B) Let the line be $L: \frac{x+1}{2} = \frac{y-3}{3} = \frac{z-1}{-1} = \lambda$. Any point $M$ on the line is $(2\lambda-1, 3\lambda+3, -\lambda+1)$

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$8$

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(B) Let the line be $L: \frac{x+1}{2} = \frac{y-3}{3} = \frac{z-1}{-1} = \lambda$. Any point $M$ on the line is $(2\lambda-1, 3\lambda+3, -\lambda+1)$.Vector $\vec{PM} = (2\lambda-1-a, 3\lambda+3-4, -\lambda+1-2) = (2\lambda-1-a, 3\lambda-1, -\lambda-1)$.Since $\vec{PM}$ is perpendicular to the line direction $\vec{v} = (2, 3, -1)$,we have $\vec{PM} \cdot \vec{v} = 0$.$2(2\lambda-1-a) + 3(3\lambda-1) - 1(-\lambda-1) = 0 \Rightarrow 4\lambda - 2 - 2a + 9\lambda - 3 + \lambda + 1 = 0 \Rightarrow 14\lambda - 4 - 2a = 0 \Rightarrow 7\lambda - 2 - a = 0 \Rightarrow a = 7\lambda - 2$.Given the length of the perpendicular $PM = 2\sqrt{6}$,so $PM^2 = 24$.$(2\lambda-1-a)^2 + (3\lambda-1)^2 + (-\lambda-1)^2 = 24$.Substituting $a = 7\lambda-2$: $(2\lambda-1-(7\lambda-2))^2 + (3\lambda-1)^2 + (\lambda+1)^2 = 24$.$(-5\lambda+1)^2 + (3\lambda-1)^2 + (\lambda+1)^2 = 24 \Rightarrow (25\lambda^2 - 10\lambda + 1) + (9\lambda^2 - 6\lambda + 1) + (\lambda^2 + 2\lambda + 1) = 24$.$35\lambda^2 - 14\lambda + 3 = 24 \Rightarrow 35\lambda^2 - 14\lambda - 21 = 0 \Rightarrow 5\lambda^2 - 2\lambda - 3 = 0$.$(5\lambda+3)(\lambda-1) = 0$. Since $a > 0$,$7\lambda-2 > 0 \Rightarrow \lambda > 2/7$. Thus $\lambda = 1$.Then $a = 7(1)-2 = 5$. Point $M = (2(1)-1, 3(1)+3, -1+1) = (1, 6, 0)$.Since $M$ is the midpoint of $PQ$,$\frac{a+\alpha_1}{2} = 1, \frac{4+\alpha_2}{2} = 6, \frac{2+\alpha_3}{2} = 0$.$\alpha_1 = 2-5 = -3, \alpha_2 = 12-4 = 8, \alpha_3 = 0-2 = -2$.$a + \alpha_1 + \alpha_2 + \alpha_3 = 5 - 3 + 8 - 2 = 8$.

If the length of the perpendicular drawn from the point $P(a, 4, 2)$,$a > 0$ on the line $\frac{x+1}{2} = \frac{y-3}{3} = \frac{z-1}{-1}$ is $2\sqrt{6}$ units and $Q(\alpha_{1}, \alpha_{2}, \alpha_{3})$ is the image of the point $P$ in this line,then $a + \sum_{i=1}^{3} \alpha_{i}$ is equal to.

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