Question

**Given:** - Line $L_1$ passes through $A = (-1,\,2,\,4)$ with direction vector $\mathbf{d_1} = (\alpha,\,-1,\,-\alpha)$. - Line $L_2$ passes through $B = (0,\,1,\,1)$ with direction vector $\mathbf{d_2} = (\alpha,\,2,\,2\alpha)$. **Formula:** $$\text{SD} = \frac{|\overrightarrow{AB} \cdot (\mathbf{d_1} \times \mathbf{d_2})|}{|\mathbf{d_1} \times \mathbf{d_2}|}$$ **Step 1:** Compute $\overrightarrow{AB} = B - A = (1,\,-1,\,-3)$. **Step 2:** Compute $\mathbf{d_1} \times \mathbf{d_2}$: $$\mathbf{d_1} \times \mathbf{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & -1 & -\alpha \\ \alpha & 2 & 2\alpha \end{vmatrix}$$ $$= \hat{i}[(-1)(2\alpha)-(-\alpha)(2)] - \hat{j}[(\alpha)(2\alpha)-(-\alpha)(\alpha)] + \hat{k}[(\alpha)(2)-(-1)(\alpha)]$$ $$= \hat{i}[-2\alpha+2\alpha] - \hat{j}[2\alpha^2+\alpha^2] + \hat{k}[2\alpha+\alpha]$$ $$= (0,\,-3\alpha^2,\,3\alpha)$$ **Step 3:** Compute the magnitude: $$|\mathbf{d_1} \times \mathbf{d_2}| = \sqrt{0 + 9\alpha^4 + 9\alpha^2} = 3|\alpha|\sqrt{\alpha^2+1}$$ **Step 4:** Compute the dot product: $$\overrightarrow{AB} \cdot (\mathbf{d_1} \times \mathbf{d_2}) = (1)(0)+(-1)(-3\alpha^2)+(-3)(3\alpha) = 3\alpha^2 - 9\alpha$$ **Step 5:** Apply the formula: $$\text{SD} = \frac{|3\alpha^2 - 9\alpha|}{3|\alpha|\sqrt{\alpha^2+1}} = \frac{3|\alpha||\alpha-3|}{3|\alpha|\sqrt{\alpha^2+1}} = \frac{|\alpha-3|}{\sqrt{\alpha^2+1}}$$ **Step 6:** Set $\text{SD} = \sqrt{2}$: $$\frac{|\alpha-3|}{\sqrt{\alpha^2+1}} = \sqrt{2}$$ Squaring both sides: $$(\alpha-3)^2 = 2(\alpha^2+1)$$ $$\alpha^2 - 6\alpha + 9 = 2\alpha^2 + 2$$ $$\alpha^2 + 6\alpha - 7 = 0$$ $$(\alpha+7)(\alpha-1) = 0$$ $$\alpha = -7 \quad \text{or} \quad \alpha = 1$$ Both values give non-zero $\mathbf{d_1} \times \mathbf{d_2}$, so both are valid. **Step 7:** Sum of all values: $$(-7) + 1 = -6$$ **Answer: (B)**