JEE Main 2023 — Mathematics Vectors & 3D Geometry

Given equations are

$\frac{x-\lambda }{3}=\frac{y-2\sqrt{6}}{4}=\frac{z+2\sqrt{6}}{5}$3x−λ​=4y−26​​=5z+26​​ is passing through a point $(\lambda ,2\sqrt{6},3\sqrt{6})$(λ,26​,36​) and it's direction ratios are $3,4,5$3,4,5.

So,

$${\vec{a}}_{1}=\lambda \hat{i}+2\sqrt{6}\hat{j}-2\sqrt{6}\hat{k}$$a1​=λi^+26​j^​−26​k^

$${\vec{b}}_{1}=3\hat{i}+4\hat{j}+5\hat{k}$$b1​=3i^+4j^​+5k^

And,

$\frac{x+\sqrt{6}}{2}=\frac{y-\sqrt{6}}{3}=\frac{z-\sqrt{6}}{4}$2x+6​​=3y−6​​=4z−6​​ is passing through a point $(-\sqrt{6},\sqrt{6},\sqrt{6})$(−6​,6​,6​) and it's direction ratios are $1,2,3$1,2,3.

$${\vec{a}}_{2}=-\sqrt{6}\hat{i}+\sqrt{6}\hat{j}+\sqrt{6}\hat{k}$$a2​=−6​i^+6​j^​+6​k^

$${\vec{b}}_{2}=2\hat{i}+3\hat{j}+4\hat{k}$$b2​=2i^+3j^​+4k^

Now,

$${\vec{a}}_{2}-{\vec{a}}_{1}=(-\sqrt{6}\hat{i}+\sqrt{6}\hat{j}+\sqrt{6}\hat{k})-(\lambda \hat{i}+2\sqrt{6}\hat{j}-2\sqrt{6}\hat{k})=-(\sqrt{6}+\lambda )\hat{i}-\sqrt{6}\hat{j}+3\sqrt{6}\hat{k}$$a2​−a1​=(−6​i^+6​j^​+6​k^)−(λi^+26​j^​−26​k^)=−(6​+λ)i^−6​j^​+36​k^

$${\vec{b}}_{1}\times {\vec{b}}_{2}=|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 5 \\ 2 & 3 & 4\end{matrix}|$$b1​×b2​=∣i^32​j^​43​k^54​∣

$$\Rightarrow {\vec{b}}_{1}\times {\vec{b}}_{2}=(\hat{i}-2\hat{j}+\hat{k})$$⇒b1​×b2​=(i^−2j^​+k^)

So,

$$\Rightarrow |{\vec{b}}_{1}\times {\vec{b}}_{2}|=\sqrt{6}$$⇒∣b1​×b2​∣=6​

We know that the shortest distance between two skew lines is $\frac{|({\vec{a}}_{2}-{\vec{a}}_{1})\cdot ({\vec{b}}_{1}\times {\vec{b}}_{2})|}{|{\vec{b}}_{1}\times {\vec{b}}_{2}|}=6$∣b1​×b2​∣∣(a2​−a1​)⋅(b1​×b2​)∣​=6

$$\Rightarrow |\frac{((-\sqrt{6}-\lambda )\hat{i}-\sqrt{6}\hat{j}+3\sqrt{6}\hat{k})\cdot (\hat{i}-2\hat{j}+\hat{k})}{\sqrt{6}}|=6$$⇒∣6​((−6​−λ)i^−6​j^​+36​k^)⋅(i^−2j^​+k^)​∣=6

$$\Rightarrow |\frac{4\sqrt{6}-\lambda }{\sqrt{6}}|=6$$⇒∣6​46​−λ​∣=6

$$\Rightarrow \lambda =10\sqrt{6},-2\sqrt{6}$$⇒λ=106​,−26​

So, required sum is

$$={(10\sqrt{6}-2\sqrt{6})}^{2}=384$$=(106​−26​)2=384