The sum of all values of alpha , for which t | vedclass

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(B) The shortest distance $d$ between two lines $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac

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-$6$

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(B) The shortest distance $d$ between two lines $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$ is given by $d = \frac{|(\vec{r_2}-\vec{r_1}) \cdot (\vec{v_1} 	imes \vec{v_2})|}{ |\vec{v_1} 	imes \vec{v_2}| }$.Here, $\vec{r_1} = (-1, 2, 4)$, $\vec{r_2} = (0, 1, 1)$, $\vec{v_1} = (\alpha, -1, -\alpha)$, and $\vec{v_2} = (\alpha, 2, 2\alpha)$.$\vec{r_2}-\vec{r_1} = (1, -1, -3)$.$\vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ \alpha & -1 & -\alpha \ \alpha & 2 & 2\alpha \end{vmatrix} = \hat{i}(-2\alpha + 2\alpha) - \hat{j}(2\alpha^2 + \alpha^2) + \hat{k}(2\alpha + \alpha) = (0, -3\alpha^2, 3\alpha)$.$|\vec{v_1} 	imes \vec{v_2}| = \sqrt{0^2 + (-3\alpha^2)^2 + (3\alpha)^2} = \sqrt{9\alpha^4 + 9\alpha^2} = 3|\alpha|\sqrt{\alpha^2+1}$.$(\vec{r_2}-\vec{r_1}) \cdot (\vec{v_1} 	imes \vec{v_2}) = (1)(0) + (-1)(-3\alpha^2) + (-3)(3\alpha) = 3\alpha^2 - 9\alpha$.Given $d = \sqrt{2}$, so $\sqrt{2} = \frac{|3\alpha^2 - 9\alpha|}{3|\alpha|\sqrt{\alpha^2+1}} = \frac{|\alpha^2 - 3\alpha|}{|\alpha|\sqrt{\alpha^2+1}} = \frac{|\alpha - 3|}{\sqrt{\alpha^2+1}}$.Squaring both sides: $2 = \frac{(\alpha-3)^2}{\alpha^2+1} \Rightarrow 2\alpha^2 + 2 = \alpha^2 - 6\alpha + 9$.$\alpha^2 + 6\alpha - 7 = 0 \Rightarrow (\alpha+7)(\alpha-1) = 0$.The values are $\alpha = -7$ and $\alpha = 1$.The sum of values is $-7 + 1 = -6$.

The sum of all values of $ \alpha $, for which the shortest distance between the lines $ \frac{x+1}{\alpha}=\frac{y-2}{-1}=\frac{z-4}{-\alpha} $ and $ \frac{x}{\alpha}=\frac{y-1}{2}=\frac{z-1}{2\alpha} $ is $ \sqrt{2} $, is

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