Animated Solution for Mathematics - Three Dimensional Geometry: If the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{1}$ and $\frac{x-a}{2}=\frac{y+2}{3}=\frac{z-3}{1}$ intersects at the point $P$, then the distance of the point $P$ from the plane $z=a$ is :

Imagine you are standing in a vast, empty room. Two laser beams are firing across the space, crossing each other at a single, precise point. In the world of 3D geometry, this is the moment of intersection.

Our goal is to find that point $P$ where two lines, $L_1$ and $L_2$, meet, and then measure its distance from a specific plane. This is the art of finding order in three-dimensional chaos.

To talk about a line in 3D, we use parameters. For the first line, $L_1:\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{1}$, we set everything equal to a parameter $\lambda$. This allows us to express any point on the line as a function of $\lambda$:

Similarly, for the second line, $L_2:\frac{x-a}{2}=\frac{y+2}{3}=\frac{z-3}{1}$, we introduce a different parameter, $\mu$. We use a different parameter because the lines do not necessarily reach the intersection point at the same "rate." Thus, we get:

At the point of intersection $P$, the coordinates must be identical. This gives us a system of equations. We start by equating the $z$-coordinates:

This serves as our bridge between the two lines. Next, we equate the $y$-coordinates: $2\lambda +2=3\mu -2$.

By substituting our bridge equation $\lambda =\mu +6$ into this, we get:

Expanding this, we find $2\mu +14=3\mu -2$, which elegantly reveals $\mu =16$. With $\mu$ in hand, $\lambda$ follows immediately: $\lambda =16+6=22$.

Now that we know exactly where the lines meet, we look at the $x$-coordinates: $\lambda +1=2\mu +a$. Substituting our values $\lambda =22$ and $\mu =16$, we get:

This leads us to $23=32+a$, or $a=-9$. We have successfully decoded the geometry of the system.

With $a=-9$, our plane is defined by $z=-9$. We calculate the intersection point $P$ by plugging $\lambda =22$ back into our $L_1$ equations:

The point $P$ is $(23,46,19)$. The distance from this point to the plane $z=-9$ is the vertical gap between the $z$-coordinate of the point and the plane itself:

We have traversed the lines, solved for the hidden constant, and measured the distance. The final distance is $28$ units.