Animated Solution for Mathematics - Three Dimensional Geometry: Let the shortest distance between the lines $L:\frac{x-5}{-2}=\frac{y-\lambda }{0}=\frac{z+\lambda }{1},\lambda \ge 0$ and $L_1:x+1=y-1=4-z$ be $2\sqrt{6}$. If $(\alpha ,\beta ,\gamma )$ lies on $L$, then which of the following is NOT possible?

A Journey into 3D Space

Welcome, my dear student! Today, we are going to peel back the layers of a beautiful 3D geometry problem. Imagine you are standing in a vast, empty room where two lines are floating in space, not parallel, and never touching.

These are what we call 'skew lines.' Our mission is to find the shortest distance between them and then explore the nature of a general point on one of these lines. This is not just algebra; it is the art of visualizing space.

The Setup and the Hidden Trap

We are given two lines, $L$ and $L_1$. Line $L$ is defined as:

Line $L_1$ is given as $x+1=y-1=4-z$. Now, pause. Do you see the trap? In JEE Advanced, the examiners love to hide small details.

Look at the $z$-term in $L_1$: $4-z$. To use our standard formulas, we must have the coefficient of the variable as $+1$. So, we rewrite $4-z$ as $-(z-4)$, which gives us:

Now, the direction vector ${\vec{b}}_2$ is clearly $\hat{i}+\hat{j}-\hat{k}$. If you missed that negative sign, the entire cross product would have been wrong. Always standardize your lines first!

The Shortest Distance

To find the shortest distance $d$ between these skew lines, we use the elegant formula:

Here, ${\vec{a}}_1=(5,\lambda ,-\lambda )$ and ${\vec{a}}_2=(-1,1,4)$ are points on the lines. The vector ${\vec{b}}_1\times {\vec{b}}_2$ represents the common normal, the direction perpendicular to both lines.

Calculating the cross product, we get:

The magnitude is $|{\vec{b}}_1\times {\vec{b}}_2|=\sqrt{(-1{)}^2+(-1{)}^2+(-2{)}^2}=\sqrt{6}$.

Solving for the Unknown

Next, we find the vector ${\vec{a}}_2-{\vec{a}}_1=(-6,1-\lambda ,4+\lambda )$. Taking the dot product with our normal vector, we get:

Substituting this into our distance formula, we have:

This simplifies to $|\lambda +3|=12$. We get $\lambda =9$ or $\lambda =-15$. Since the problem constraints demand $\lambda \ge 0$, we confidently choose $\lambda =9$.

The Final Verification

With $\lambda =9$, line $L$ becomes:

This gives us the parametric form: $\alpha =-2k+5$, $\beta =9$, and $\gamma =k-9$. The question asks which option is NOT possible.

Let's test the first option: $\alpha +2\gamma$. Substituting our parameters, we get:

The parameter $k$ cancels out completely! This means for any point on line $L$, $\alpha +2\gamma$ must be $-13$. Since the first option claims it is $24$, it is mathematically impossible. You have just mastered the geometry of skew lines!