Animated Solution for Mathematics - Three Dimensional Geometry: If the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect, then $k$ is equal to:

Imagine you are standing in a vast, empty room. You hold two laser pointers, each casting a beam of light across the space. In the chaotic world of 3D geometry, these two beams are most likely 'skew'—they pass each other like ships in the night, never touching.

But today, we are not interested in the ordinary. We are looking for the extraordinary moment where these two beams collide. We are looking for the intersection, a search for a point of convergence.

Every line in 3D space carries its identity in its equation. When we look at $L_1:\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$, we aren't just seeing numbers; we are seeing a path.

We extract the point $A(1,-1,1)$ and the direction vector $\overrightarrow{b_1}=2\hat{i}+3\hat{j}+4\hat{k}$. Similarly, for $L_2:\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$, we identify point $B(3,k,0)$ and direction vector $\overrightarrow{b_2}=1\hat{i}+2\hat{j}+1\hat{k}$.

This is the DNA of our lines. Without these, we are blind; with them, we are architects of the space.

Here is the secret that separates the masters from the novices. For two lines to intersect, they must be coplanar. They must lie on the same flat sheet of paper.

If they are coplanar, then the vector connecting any point on the first line to any point on the second line—let us call it $\overrightarrow{AB}$—must lie in the same plane as the direction vectors $\overrightarrow{b_1}$ and $\overrightarrow{b_2}$. Mathematically, this means the scalar triple product of these three vectors must vanish.

The scalar triple product represents the volume of a parallelepiped. If the volume is zero, the parallelepiped has collapsed into a plane, and the lines are locked together.

Let us construct our vector $\overrightarrow{AB}$. By subtracting the coordinates of $A$ from $B$, we get $\overrightarrow{AB}=(3-1)\hat{i}+(k-(-1))\hat{j}+(0-1)\hat{k}$, which simplifies to $\overrightarrow{AB}=2\hat{i}+(k+1)\hat{j}-1\hat{k}$.

Now, we assemble the determinant:

We expand along the first row:

Let us breathe and simplify. The first term becomes $2(3-8)=-10$. The second term becomes $-(k+1)(2-4)=-(k+1)(-2)=2(k+1)$.

The third term becomes $-1(4-3)=-1$. Putting it all together:

This simplifies beautifully to $2k-9=0$. Solving for $k$, we find:

The lines have met. The mystery is solved. You have successfully navigated the 3D landscape, proving that with the right tools—the scalar triple product and the determinant—even the most complex spatial problems yield to your intellect.