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 :

3D Geometry PYQ — JEE Main 2023 · 24 Jan · Shift 2 · Q38 with Solution

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Correct answer: 384

The given lines are

$\frac{x+\sqrt6}{2}=\frac{y-\sqrt6}{3}=\frac{z-\sqrt6}{4}$

and

$\frac{x-\lambda}{3}=\frac{y-2\sqrt6}{4}=\frac{z+2\sqrt6}{5}$

So their parametric forms are:

Line $L_1$ : $x=-\sqrt6+2t,\quad y=\sqrt6+3t,\quad z=\sqrt6+4t$ Hence a point on $L_1$ is $A(-\sqrt6,\sqrt6,\sqrt6)$ and direction vector is $\vec d_1=(2,3,4).$
Line $L_2$ : $x=\lambda+3s,\quad y=2\sqrt6+4s,\quad z=-2\sqrt6+5s$ Hence a point on $L_2$ is $B(\lambda,2\sqrt6,-2\sqrt6)$ and direction vector is $\vec d_2=(3,4,5).$

For lines through points $A,B$ with direction vectors $\vec d_1,\vec d_2$ ,

$\text{Shortest distance} = \frac{|(\overrightarrow{AB})\cdot(\vec d_1\times \vec d_2)|}{|\vec d_1\times \vec d_2|}.$

Here,

$\overrightarrow{AB}=B-A=(\lambda+\sqrt6,\sqrt6,-3\sqrt6).$

\begin{vmatrix} \hat i & \hat j & \hat k \\ 2&3&4\\ 3&4&5 \end{vmatrix}$$ $$=\hat i(15-16)-\hat j(10-12)+\hat k(8-9)$$ $$=(-1,2,-1).$$ Its magnitude is $$|\vec d_1\times \vec d_2|=\sqrt{(-1)^2+2^2+(-1)^2}=\sqrt6.$$ --- 4. **Compute the scalar triple product** $$\overrightarrow{AB}\cdot(\vec d_1\times\vec d_2) =(\lambda+\sqrt6,\sqrt6,-3\sqrt6)\cdot(-1,2,-1).$$ So, $$=-(\lambda+\sqrt6)+2\sqrt6+3\sqrt6=-\lambda+4\sqrt6.$$ Therefore shortest distance is $$\frac{|-lambda+4\sqrt6|}{\sqrt6}=6.$$ That is, $$|-lambda+4\sqrt6|=6\sqrt6.$$ So, $$-\lambda+4\sqrt6=\pm 6\sqrt6.$$ Hence the possible values of $\lambda$ are: - If $-\lambda+4\sqrt6=6\sqrt6$, then $$-\lambda=2\sqrt6\implies \lambda=-2\sqrt6.$$ - If $-\lambda+4\sqrt6=-6\sqrt6$, then $$-\lambda=-10\sqrt6\implies \lambda=10\sqrt6.$$ Thus, $$\lambda_1=-2\sqrt6,\qquad \lambda_2=10\sqrt6.$$ --- 5. **Find the square of the sum of all possible values** Sum of all possible values: $$\lambda_1+\lambda_2=-2\sqrt6+10\sqrt6=8\sqrt6.$$ Required square: $$(8\sqrt6)^2=64\cdot 6=384.$$ --- 6. **Compare with stored answer** Derived answer is $$384$$ which matches the stored correct answer.