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

The problem involves finding the value of $$\alpha$$ such that the shortest distance between the given line $$\frac{x-1}{\alpha} = \frac{y+1}{-1} = \frac{z}{1}$$ (with $$\alpha \neq -1$$) and the line of intersection of the planes $$x + y + z + 1 = 0$$ and $$2x - y + z + 3 = 0$$ is $$\frac{1}{\sqrt{3}}$$. To solve this, we first find the direction vector and a point on the line of intersection of the two planes. The normal vector to the first plane $$x + y + z + 1 = 0$$ is $$\langle 1, 1, 1 \rangle$$, and to the second plane $$2x - y + z + 3 = 0$$ is $$\langle 2, -1, 1 \rangle$$. The direction vector of the line of intersection is the cross product of these normal vectors: $$ \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & -1 & 1 \\ \end{vmatrix} = \hat{i}(1 \cdot 1 - 1 \cdot (-1)) - \hat{j}(1 \cdot 1 - 1 \cdot 2) + \hat{k}(1 \cdot (-1) - 1 \cdot 2) = \hat{i}(1 + 1) - \hat{j}(1 - 2) + \hat{k}(-1 - 2) = \langle 2, 1, -3 \rangle. $$ To find a point on this line, set $$z = 0$$ and solve the system: $$$ x + y + 1 = 0 \quad \Rightarrow \quad x + y = -1, $$$ $$$ 2x - y + 3 = 0 \quad \Rightarrow \quad 2x - y = -3. $$$ Adding these equations: $$3x = -4$$ gives $$x = -\frac{4}{3}$$. Substituting into $$x + y = -1$$: $$-\frac{4}{3} + y = -1$$ gives $$y = -1 + \frac{4}{3} = \frac{1}{3}$$. So, a point is $$\left( -\frac{4}{3}, \frac{1}{3}, 0 \right)$$, denoted as $$\vec{a_2} = \langle -\frac{4}{3}, \frac{1}{3}, 0 \rangle$$. The given line $$\frac{x-1}{\alpha} = \frac{y+1}{-1} = \frac{z}{1}$$ can be parameterized by setting it equal to $$\lambda$$: $$$ x = 1 + \alpha \lambda, \quad y = -1 - \lambda, \quad z = \lambda. $$$ When $$\lambda = 0$$, a point is $$(1, -1, 0)$$, so $$\vec{a_1} = \langle 1, -1, 0 \rangle$$. The direction vector is $$\vec{b_1} = \langle \alpha, -1, 1 \rangle$$. The vector between the points is: $$$ \vec{a_2} - \vec{a_1} = \left\langle -\frac{4}{3} - 1, \frac{1}{3} - (-1), 0 - 0 \right\rangle = \left\langle -\frac{7}{3}, \frac{4}{3}, 0 \right\rangle. $$$ The shortest distance $$d$$ between two skew lines is given by: $$$ d = \frac{| (\vec{b_1} \times \vec{b_2}) \cdot (\vec{a_2} - \vec{a_1}) |}{|\vec{b_1} \times \vec{b_2}|}. $$$ First, compute $$\vec{b_1} \times \vec{b_2}$$: $$$ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & -1 & 1 \\ 2 & 1 & -3 \\ \end{vmatrix} = \hat{i}[(-1)(-3) - (1)(1)] - \hat{j}[(\alpha)(-3) - (1)(2)] + \hat{k}[(\alpha)(1) - (-1)(2)] = \hat{i}(3 - 1) - \hat{j}(-3\alpha - 2) + \hat{k}(\alpha + 2) = \langle 2, 3\alpha + 2, \alpha + 2 \rangle. $$$ Now, compute the dot product with $$\vec{a_2} - \vec{a_1}$$: $$$ (\vec{b_1} \times \vec{b_2}) \cdot \left\langle -\frac{7}{3}, \frac{4}{3}, 0 \right\rangle = (2)\left(-\frac{7}{3}\right) + (3\alpha + 2)\left(\frac{4}{3}\right) + (\alpha + 2)(0) = -\frac{14}{3} + \frac{4}{3}(3\alpha + 2) = -\frac{14}{3} + 4\alpha + \frac{8}{3} = 4\alpha - \frac{6}{3} = 4\alpha - 2. $$$ The absolute value is $$|4\alpha - 2|$$. Next, compute $$|\vec{b_1} \times \vec{b_2}|$$: $$$ |\vec{b_1} \times \vec{b_2}| = \sqrt{(2)^2 + (3\alpha + 2)^2 + (\alpha + 2)^2} = \sqrt{4 + (9\alpha^2 + 12\alpha + 4) + (\alpha^2 + 4\alpha + 4)} = \sqrt{10\alpha^2 + 16\alpha + 12}. $$$ Given that the distance is $$\frac{1}{\sqrt{3}}$$: $$$ \frac{|4\alpha - 2|}{\sqrt{10\alpha^2 + 16\alpha + 12}} = \frac{1}{\sqrt{3}}. $$$ Rewriting $$|4\alpha - 2| = 2|2\alpha - 1|$$: $$$ \frac{2|2\alpha - 1|}{\sqrt{10\alpha^2 + 16\alpha + 12}} = \frac{1}{\sqrt{3}}. $$$ Multiply both sides by $$\sqrt{3}$$: $$$ 2|2\alpha - 1| \sqrt{3} = \sqrt{10\alpha^2 + 16\alpha + 12}. $$$ Square both sides to eliminate absolute values and square roots: $$$ (2|2\alpha - 1| \sqrt{3})^2 = (\sqrt{10\alpha^2 + 16\alpha + 12})^2, $$$ $$$ 4 \cdot (2\alpha - 1)^2 \cdot 3 = 10\alpha^2 + 16\alpha + 12, $$$ $$$ 12(4\alpha^2 - 4\alpha + 1) = 10\alpha^2 + 16\alpha + 12, $$$ $$$ 48\alpha^2 - 48\alpha + 12 = 10\alpha^2 + 16\alpha + 12. $$$ Bring all terms to one side: $$$ 48\alpha^2 - 48\alpha + 12 - 10\alpha^2 - 16\alpha - 12 = 0, $$$ $$$ 38\alpha^2 - 64\alpha = 0, $$$ $$$ 2\alpha(19\alpha - 32) = 0. $$$ So, $$\alpha = 0$$ or $$\alpha = \frac{32}{19}$$. Now, check these values in the original distance equation. For $$\alpha = 0$$: $$$ |4(0) - 2| = |-2| = 2, \quad \sqrt{10(0)^2 + 16(0) + 12} = \sqrt{12} = 2\sqrt{3}, $$$ $$$ d = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}. $$$ However, when $$\alpha = 0$$, the symmetric equation of the line $$\frac{x-1}{0} = \frac{y+1}{-1} = \frac{z}{1}$$ is undefined (division by zero). Thus, $$\alpha = 0$$ is invalid. For $$\alpha = \frac{32}{19}$$: $$$ |4\left(\frac{32}{19}\right) - 2| = \left|\frac{128}{19} - \frac{38}{19}\right| = \left|\frac{90}{19}\right| = \frac{90}{19}, $$$ $$$ \sqrt{10\left(\frac{32}{19}\right)^2 + 16\left(\frac{32}{19}\right) + 12} = \sqrt{\frac{10240}{361} + \frac{512}{19} + 12} = \sqrt{\frac{10240}{361} + \frac{512 \cdot 19}{361} + \frac{12 \cdot 361}{361}} = \sqrt{\frac{10240 + 9728 + 4332}{361}} = \sqrt{\frac{24300}{361}} = \frac{\sqrt{24300}}{19} = \frac{90\sqrt{3}}{19}, $$$ $$$ d = \frac{\frac{90}{19}}{\frac{90\sqrt{3}}{19}} = \frac{90}{19} \cdot \frac{19}{90\sqrt{3}} = \frac{1}{\sqrt{3}}. $$$ This satisfies the distance condition. Also, $$\alpha = \frac{32}{19} \neq -1$$, and the lines are skew (direction vectors $$\langle \frac{32}{19}, -1, 1 \rangle$$ and $$\langle 2, 1, -3 \rangle$$ are not parallel since $$\frac{32/19}{2} = \frac{32}{38} = \frac{16}{19}$$, $$\frac{-1}{1} = -1$$, $$\frac{1}{-3} = -\frac{1}{3}$$, and these ratios are unequal). Comparing with the options: - A. $$-\frac{19}{16}$$ - B. $$\frac{32}{19}$$ - C. $$-\frac{16}{19}$$ - D. $$\frac{19}{32}$$ The value $$\alpha = \frac{32}{19}$$ corresponds to option B. Hence, the correct answer is Option B.