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(A) The shortest distance 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} = \

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$16$

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(A) The shortest distance 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 $d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 	imes \vec{b}_2)|}{|\vec{b}_1 	imes \vec{b}_2|}$.Here,$\vec{a}_1 = (1, 2, 3)$,$\vec{a}_2 = (2, 4, 5)$,$\vec{b}_1 = (2, 3, \lambda)$,and $\vec{b}_2 = (1, 4, 5)$.$\vec{b}_1 	imes \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & \lambda \ 1 & 4 & 5 \end{vmatrix} = \hat{i}(15-4\lambda) - \hat{j}(10-\lambda) + \hat{k}(8-3) = (15-4\lambda)\hat{i} + (\lambda-10)\hat{j} + 5\hat{k}$.$\vec{a}_2 - \vec{a}_1 = (2-1)\hat{i} + (4-2)\hat{j} + (5-3)\hat{k} = \hat{i} + 2\hat{j} + 2\hat{k}$.$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 	imes \vec{b}_2) = (1)(15-4\lambda) + 2(\lambda-10) + 2(5) = 15 - 4\lambda + 2\lambda - 20 + 10 = 5 - 2\lambda$.$|\vec{b}_1 	imes \vec{b}_2| = \sqrt{(15-4\lambda)^2 + (\lambda-10)^2 + 25} = \sqrt{225 + 16\lambda^2 - 120\lambda + \lambda^2 - 20\lambda + 100 + 25} = \sqrt{17\lambda^2 - 140\lambda + 350}$.Given $d = \frac{1}{\sqrt{3}}$,so $\frac{|5-2\lambda|}{\sqrt{17\lambda^2 - 140\lambda + 350}} = \frac{1}{\sqrt{3}}$.Squaring both sides: $3(5-2\lambda)^2 = 17\lambda^2 - 140\lambda + 350$.$3(25 - 20\lambda + 4\lambda^2) = 17\lambda^2 - 140\lambda + 350 \implies 75 - 60\lambda + 12\lambda^2 = 17\lambda^2 - 140\lambda + 350$.$5\lambda^2 - 80\lambda + 275 = 0 \implies \lambda^2 - 16\lambda + 55 = 0$.$(\lambda-5)(\lambda-11) = 0$,so $\lambda = 5$ or $\lambda = 11$.The sum of all possible values of $\lambda$ is $5 + 11 = 16$.

If the shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{\lambda}$ and $\frac{x-2}{1}=\frac{y-4}{4}=\frac{z-5}{5}$ is $\frac{1}{\sqrt{3}}$,then the sum of all possible values of $\lambda$ is

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