QJEEM23Ma81f7: JEE Main 2023 Mathematics - Class 12 JEEM23Ma81f7 Solution: Let be the set of all values of , for which the shortest…

Given the two lines :

$\frac{x-\lambda}{0}=\frac{y-3}{4}=\frac{z+6}{1} \\\\\frac{x+\lambda}{3}=\frac{y}{-4}=\frac{z-6}{0}$

Let's find the direction vectors of these lines: $\vec{d_1} = \langle 0, 4, 1 \rangle$ and $\vec{d_2} = \langle 3, -4, 0 \rangle$ .

Now, let's find the cross product of the direction vectors, which will give a vector that is perpendicular to both lines :

$\vec{n} = \vec{d_1} \times \vec{d_2} = \langle 4, 3, -12 \rangle$

Let's find the vector connecting a point on line 1 to a point on line 2 :

$\vec{c} = \langle 2\lambda, 3, -12 \rangle$

The shortest distance between the two lines is the projection of $\vec{c}$ onto $\vec{n}$ :

$d = \left|\frac{\vec{c} \cdot \vec{n}}{|\vec{n}|}\right| = \left|\frac{(2\lambda)(4) + (3)(3) - (12)(-12)}{\sqrt{16 + 9 + 144}}\right|$

We are given that the shortest distance is 13 :

$13 = \left|\frac{8\lambda + 153}{13}\right|$

$|8\lambda + 153| = 169$

We have two cases :

1. $8\lambda + 153 = 169$

$\lambda = \frac{16}{8}$

2. $8\lambda + 153 = -169$

$\lambda = \frac{-322}{8}$

Now, let's calculate $8\left|\sum\limits_{\lambda \in S} \lambda\right|$ :

$8\left|\frac{16}{8} + \frac{-322}{8}\right| = 8\left|\frac{-306}{8}\right| = 306$

Thus, the correct answer is Option A : 306.