JEE MAIN - Mathematics (2023 - 24th January Evening Shift - No. 16)
If the shortest between the lines $${{x + \sqrt 6 } \over 2} = {{y - \sqrt 6 } \over 3} = {{z - \sqrt 6 } \over 4}$$ and $${{x - \lambda } \over 3} = {{y - 2\sqrt 6 } \over 4} = {{z + 2\sqrt 6 } \over 5}$$ is 6, then the square of sum of all possible values of $$\lambda$$ is :
Answer
384
Explanation
Shortest distance between the lines
$\frac{x+\sqrt{6}}{2}=\frac{y-\sqrt{6}}{3}=\frac{z-\sqrt{6}}{4}$
$\frac{x-\lambda}{3}=\frac{y-2 \sqrt{6}}{4}=\frac{2+2 \sqrt{6}}{5}$ is 6
Vector along line of shortest distance
$=\left|\begin{array}{lll}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5\end{array}\right|, \Rightarrow-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\mathrm{k}$ (its magnitude is $\sqrt{6}$ )
Now $\frac{1}{\sqrt{6}}\left|\begin{array}{ccc}\sqrt{6}+\lambda & \sqrt{6} & -3 \sqrt{6} \\ 2 & 3 & 4 \\ 3 & 4 & 5\end{array}\right|=\pm 6$
$\Rightarrow \lambda=-2 \sqrt{6}, 10 \sqrt{6}$
So, square of sum of these values
$$ =(10 \sqrt{6}-2 \sqrt{6})^2=(8 \sqrt{6})^2=384 $$
$\frac{x+\sqrt{6}}{2}=\frac{y-\sqrt{6}}{3}=\frac{z-\sqrt{6}}{4}$
$\frac{x-\lambda}{3}=\frac{y-2 \sqrt{6}}{4}=\frac{2+2 \sqrt{6}}{5}$ is 6
Vector along line of shortest distance
$=\left|\begin{array}{lll}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5\end{array}\right|, \Rightarrow-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\mathrm{k}$ (its magnitude is $\sqrt{6}$ )
Now $\frac{1}{\sqrt{6}}\left|\begin{array}{ccc}\sqrt{6}+\lambda & \sqrt{6} & -3 \sqrt{6} \\ 2 & 3 & 4 \\ 3 & 4 & 5\end{array}\right|=\pm 6$
$\Rightarrow \lambda=-2 \sqrt{6}, 10 \sqrt{6}$
So, square of sum of these values
$$ =(10 \sqrt{6}-2 \sqrt{6})^2=(8 \sqrt{6})^2=384 $$
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