### Three Dimensional Geometry Question 1

#### Question 1 - 2024 (01 Feb Shift 1)

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

(1) 0

(2) $2 \sqrt{3}$

(3) $3 \sqrt{3}$

(4) $-2 \sqrt{3}$

## Show Answer

#### Answer (2)

#### Solution

Passing points of lines $L _1 \& L _2$ are $(\lambda, 2,1) \&(\sqrt{3}, 1,2)$

$S . D=\frac{\left|\begin{array}{ccc}\sqrt{3}-\lambda & -1 & 1 \\ -2 & 1 & 1 \\ 1 & -2 & 1\end{array}\right|}{\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ -2 & 1 & 1 \\ 1 & -2 & 1\end{array}\right| \text { th }}$

$1=\left|\frac{\sqrt{3}-\lambda}{\sqrt{3}}\right|$

$\lambda=0, \lambda=2 \sqrt{3}$