JEE Main 2023 — Mathematics Vectors & 3D Geometry

The given lines are$\frac{x-\lambda }{0}=\frac{y-3}{4}=\frac{z+6}{1}$0x−λ​=4y−3​=1z+6​and $\frac{x+\lambda }{3}=\frac{y}{-4}=\frac{z-6}{0}$3x+λ​=−4y​=0z−6​

We know that the shortest distance between two lines is d=|\frac{({\vec{a}}_{2}-{\vec{a}}_{1})\cdot ({\vec{n}}_{1}\times {\vec{n}}_{2})}{|{\vec{n}}_{1}\times {\vec{n}}_{2}|}|&d=13(\text{given})

$\Rightarrow {\vec{n}}_{1}\times {\vec{n}}_{2}=|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 0 & 4 & 1 \\ 3 & -4 & 0\end{matrix}|$⇒n1​×n2​=∣i^03​j^​4−4​k^10​∣$=4\hat{i}+3\hat{j}-12\hat{k}$=4i^+3j^​−12k^

Now using the above formula of distance we get,

$$\Rightarrow |\frac{(2\lambda \hat{i}+3\hat{j}-12\hat{k})\cdot (4\hat{i}+3\hat{j}-12\hat{k})}{\sqrt{16+9+144}}|=13$$⇒∣16+9+144​(2λi^+3j^​−12k^)⋅(4i^+3j^​−12k^)​∣=13

$$\Rightarrow |\frac{(2\lambda \hat{i}+3\hat{j}-12\hat{k})\cdot (4\hat{i}+3\hat{j}-12\hat{k})}{13}|=13$$⇒∣13(2λi^+3j^​−12k^)⋅(4i^+3j^​−12k^)​∣=13

$$\Rightarrow |8\lambda +9+144|=169$$⇒∣8λ+9+144∣=169

$$\Rightarrow |8\lambda +153|=169$$⇒∣8λ+153∣=169

$$\Rightarrow 8\lambda =\pm 169-153$$⇒8λ=±169−153

$$\Rightarrow \lambda =\frac{16}{8},\frac{-322}{8}$$⇒λ=816​,8−322​

$$\Rightarrow 8|\underset{\lambda \in s}{\sum }\lambda |=8|\frac{16}{8}-\frac{322}{8}|=306$$⇒8∣λ∈s∑​λ∣=8∣816​−8322​∣=306

Hence this is the correct option.