If the shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and…

If the shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x}{1}=\frac{y}{\alpha}=\frac{z-5}{1}$ is $\frac{5}{\sqrt{6}}$, then the sum of all possible values of $\alpha$ is
  1. $\frac{3}{2}$
  2. $-\frac{3}{2}$
  3. 3
  4. -3

Solution


$\begin{aligned} & L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} \\ & L_1: \frac{x}{1}=\frac{y}{\alpha}=\frac{z-5}{1} \\ & \vec{x}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 1 & \alpha & 1\end{array}\right|=\hat{\mathrm{i}}(3-4 \alpha)-\hat{\mathrm{j}}(-2)+\hat{\mathrm{k}}(2 \alpha-3) \\ & \text { S.D. }=\left|\frac{\overrightarrow{\mathrm{BA}} \cdot \overrightarrow{\mathrm{n}}}{|\overrightarrow{\mathrm{n}}|}\right|=\left|\frac{(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}) \cdot \vec{n}}{|\overrightarrow{\mathrm{n}}|}\right| \\ & \Rightarrow 6(13-8 \alpha)^2=25\left((4 \alpha-3)^2+(2 \alpha-3)^2+16\right) \\ & 6\left(64 a^2-280 \alpha+169\right)=25\left(20 \alpha^2-36 \alpha+34\right) \\ & \Rightarrow 116 \alpha^2+348 \alpha-164=0 \\ & \alpha_1+\alpha_2=\frac{-348}{116}=-3\end{aligned}$ .

Asked in: JEE Main 2025 (07 Apr Shift 1)

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