Q.Check whether the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 4}{5} = \frac{y - 1}{2} = z$ are parallel or not. If they are parallel, find the distance between them, otherwise if the lines are intersecting, find their point of intersection.

The key idea is to determine the relative position of two lines in 3D space: parallel, intersecting, or skew.

1. Extract Direction Vectors:

   For line $L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$, the direction vector is $\vec{b_1} = (2, 3, 4)$.

   For line $L_2: \frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z}{1}$, the direction vector is $\vec{b_2} = (5, 2, 1)$.

2. Check for Parallelism:

   The lines are parallel if their direction vectors are proportional.

   $\frac{2}{5} \neq \frac{3}{2} \neq \frac{4}{1}$.

   Since the direction ratios are not proportional, the lines are not parallel.

3. Check for Intersection:

   Since the lines are not parallel, they either intersect or are skew. To check for intersection, we equate the general points on each line.

   Let $P_1(\lambda) = (2\lambda + 1, 3\lambda + 2, 4\lambda + 3)$ be a point on $L_1$.

   Let $P_2(\mu) = (5\mu + 4, 2\mu + 1, \mu)$ be a point on $L_2$.

   If they intersect, $P_1(\lambda) = P_2(\mu)$:

   $2\lambda + 1 = 5\mu + 4 \implies 2\lambda - 5\mu = 3 \quad (i)$

   $3\lambda + 2 = 2\mu + 1 \implies 3\lambda - 2\mu = -1 \quad (ii)$

   $4\lambda + 3 = \mu \quad (iii)$

   Substitute $(iii)$ into $(i)$:

   $2\lambda - 5(4\lambda + 3) = 3$

   $2\lambda - 20\lambda - 15 = 3$ …

When dealing with lines in three-dimensional space, there are three possibilities for their relative orientation:

1. Parallel lines: Their direction vectors are proportional, and they never meet.
2. Intersecting lines: Their direction vectors are not proportional, and they meet at exactly one point.
3. Skew lines: Their direction vectors are not proportional, and they do not meet. They lie in different planes.

Our strategy will be to first check if the lines are parallel by comparing their direction vectors. If they are not parallel, we will then attempt to find a point of intersection. If a common point exists, they are intersecting; otherwise, they are skew.

1. Represent the lines in vector form:

   The given Cartesian equations of the lines are:

   Line 1 ($L_1$): $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$

   Line 2 ($L_2$): $\frac{x - 4}{5} = \frac{y - 1}{2} = z$ (which can be written as $\frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z - 0}{1}$)

   From these, we can identify the position vectors (points through which the lines pass) and the direction vectors for each line.

   For $L_1$:

   Position vector $\vec{a_1} = \hat{i} + 2\hat{j} + 3\hat{k}$

   Direction vector $\vec{b_1} = 2\hat{i} + 3\hat{j} + 4\hat{k}$

   For $L_2$:

   Position vector $\vec{a_2} = 4\hat{i} + \hat{j} + 0\hat{k}$

   Direction vector $\vec{b_2} = 5\hat{i} + 2\hat{j} + \hat{k}$

2. Check for parallelism:

   Two lines are parallel if their direction vectors are proportional, meaning $\vec{b_1} = k \vec{b_2}$ for some scalar $k$.

   Comparing the components of $\vec{b_1}$ and $\vec{b_2}$:

   $\frac{2}{5} \neq \frac{3}{2} \neq \frac{4}{1}$

   Since the ratios of corresponding components are not equal, the direction vectors are not proportional. Therefore, the lines are not parallel.

3. Check for intersection:

   Since the lines are not parallel, they are either intersecting or skew. To check for intersection, we find the general coordinates of a point on each line using parameters, say $s$ for $L_1$ and $t$ for $L_2$, and then equate them.

   For $L_1$, let $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} = s$.

   A general point $P_1$ on $L_1$ has coordinates $(1 + 2s, 2 + 3s, 3 + 4s)$.

   For $L_2$, let $\frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z - 0}{1} = t$.

   A general point $P_2$ on $L_2$ has coordinates $(4 + 5t, 1 + 2t, t)$.

   If the lines intersect, there must be values of $s$ and $t$ such that $P_1 = P_2$. Equating the corresponding coordinates:

   1. $1 + 2s = 4 + 5t \implies 2s - 5t = 3$
   2. $2 + 3s = 1 + 2t \implies 3s - 2t = -1$ …