It seems that the video transcript provided does not contain relevant information or steps to solve the given problem regarding the lines \( L_1 \) and \( L_2 \). However, I can help you solve the problem based on the information given in the question.
Let's consider the lines:
1. \( L_1: x + y + z = 3 \)
2. \( L_2: x + 2y + 3z = 1 \)
### Step 1: Identify the equations of the lines
We have two equations representing the lines in three-dimensional space. The first line \( L_1 \) is given by the equation \( x + y + z = 3 \) and the second line \( L_2 \) is given by \( x + 2y + 3z = 1 \).
### Step 2: Find the direction ratios of the lines
To find the direction ratios of the lines, we can rewrite the equations in parametric form.
For \( L_1 \):
- Let \( z = t \) (parameter)
- Then, \( x + y = 3 - t \)
We can express \( x \) and \( y \) in terms of \( t \):
- Let \( y = s \) (another parameter)
- Then, \( x = 3 - t - s \)
Thus, the parametric equations for \( L_1 \) can be written as:
- \( x = 3 - t - s \)
- \( y = s \)
- \( z = t \)
For \( L_2 \):
- Let \( z = u \) (parameter)
- Then, \( x + 2y = 1 - 3u \)
We can express \( x \) and \( y \) in terms of \( u \):
- Let \( y = v \) (another parameter)
- Then, \( x = 1 - 3u - 2v \)
Thus, the parametric equations for \( L_2 \) can be written as:
- \( x = 1 - 3u - 2v \)
- \( y = v \)
- \( z = u \)
### Step 3: Determine if the lines are parallel or intersecting
To check if the lines are parallel, we can compare their direction ratios.
For \( L_1 \), the direction ratios can be obtained from the coefficients of \( x, y, z \):
- Direction ratios of \( L_1 \): \( (1, 1, 1) \)
For \( L_2 \), the direction ratios are:
- Direction ratios of \( L_2 \): \( (1, 2, 3) \)
Since the direction ratios are not proportional, the lines are not parallel.
### Step 4: Find the point of intersection
To find the point of intersection, we need to solve the system of equations:
1. \( x + y + z = 3 \)
2. \( x + 2y + 3z = 1 \)
We can substitute \( z = 3 - x - y \) from the first equation into the second equation:
\[
x + 2y + 3(3 - x - y) = 1
\]
Expanding this gives:
\[
x + 2y + 9 - 3x - 3y = 1
\]
Combining like terms:
\[
-2x - y + 9 = 1
\]
Rearranging gives:
\[
2x + y = 8 \quad (3)
\]
Now we have a new equation (3) along with equation (1):
1. \( x + y + z = 3 \)
2. \( 2x + y = 8 \)
We can express \( y \) from equation (3):
\[
y = 8 - 2x
\]
Substituting \( y \) back into equation (1):
\[
x + (8 - 2x) + z = 3
\]
This simplifies to:
\[
-x + 8 + z = 3
\]
Rearranging gives:
\[
z = x - 5 \quad (4)
\]
Now we can express \( x, y, z \) in terms of a single parameter \( x \):
- \( y = 8 - 2x \)
- \( z = x - 5 \)
### Step 5: Conclusion
The lines \( L_1 \) and \( L_2 \) intersect at the point \( (x, y, z) \) where \( x \) can take any value, leading to a line of intersection.
