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(C) The second line is given by the intersection of two planes: $x + y + z + 1 = 0$ and $2x - y + z + 3 = 0$. The equation of the family of planes pas

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$\frac{32}{19}$

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(C) The second line is given by the intersection of two planes: $x + y + z + 1 = 0$ and $2x - y + z + 3 = 0$. The equation of the family of planes passing through the line of intersection is $(x + y + z + 1) + \lambda(2x - y + z + 3) = 0$,which simplifies to $(1 + 2\lambda)x + (1 - \lambda)y + (1 + \lambda)z + (1 + 3\lambda) = 0$. The first line has direction vector $\vec{v_1} = (\alpha, -1, 1)$. The normal to the plane is $\vec{n} = (1 + 2\lambda, 1 - \lambda, 1 + \lambda)$. Since the line is parallel to the plane,$\vec{v_1} \cdot \vec{n} = 0$,so $\alpha(1 + 2\lambda) - (1 - \lambda) + (1 + \lambda) = 0$,which gives $\alpha(1 + 2\lambda) + 2\lambda = 0$,or $\alpha = -\frac{2\lambda}{1 + 2\lambda}$. The shortest distance between the line and the plane is the perpendicular distance from any point on the line (e.g.,$(1, -1, 0)$) to the plane: $d = \frac{|(1 + 2\lambda)(1) + (1 - \lambda)(-1) + (1 + \lambda)(0) + (1 + 3\lambda)|}{\sqrt{(1 + 2\lambda)^2 + (1 - \lambda)^2 + (1 + \lambda)^2}} = \frac{1}{\sqrt{3}}$. Simplifying the numerator: $|1 + 2\lambda - 1 + \lambda + 1 + 3\lambda| = |6\lambda + 1|$. Simplifying the denominator: $\sqrt{1 + 4\lambda + 4\lambda^2 + 1 - 2\lambda + \lambda^2 + 1 + 2\lambda + \lambda^2} = \sqrt{6\lambda^2 + 4\lambda + 3}$. Squaring both sides: $\frac{(6\lambda + 1)^2}{6\lambda^2 + 4\lambda + 3} = \frac{1}{3} \Rightarrow 3(36\lambda^2 + 12\lambda + 1) = 6\lambda^2 + 4\lambda + 3$. $108\lambda^2 + 36\lambda + 3 = 6\lambda^2 + 4\lambda + 3 \Rightarrow 102\lambda^2 + 32\lambda = 0$. Thus,$\lambda = 0$ or $\lambda = -\frac{32}{102} = -\frac{16}{51}$. If $\lambda = 0$,$\alpha = 0$. If $\lambda = -\frac{16}{51}$,$\alpha = -\frac{2(-16/51)}{1 + 2(-16/51)} = \frac{32/51}{(51 - 32)/51} = \frac{32}{19}$.

If the shortest distance between the lines $\frac{x - 1}{\alpha} = \frac{y + 1}{-1} = \frac{z}{1}, (\alpha \ne -1)$ and $x + y + z + 1 = 0 = 2x - y + z + 3$ is $\frac{1}{\sqrt{3}}$,then a value of $\alpha$ is

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