To find the shortest distance between the two lines \( L_1 \) and \( L_2 \), we will follow these steps:
### Step 1: Identify the lines in vector form
The lines are given as:
- \( L_1: x + y + z = 3 \)
- \( L_2: x + 2y + 3z = 1 \)
We can express these lines in vector form.
For \( L_1 \), we can write it as:
\[
\vec{r_1} = \vec{a_1} + \lambda \vec{b_1}
\]
where \( \vec{a_1} \) is a point on the line and \( \vec{b_1} \) is the direction vector.
For \( L_2 \), we can write it as:
\[
\vec{r_2} = \vec{a_2} + \mu \vec{b_2}
\]
where \( \vec{a_2} \) is a point on the line and \( \vec{b_2} \) is the direction vector.
### Step 2: Determine points and direction vectors
From the equations of the lines, we can find points and direction vectors.
For \( L_1 \):
- A point \( \vec{a_1} = (1, 1, 1) \) (substituting \( x=1, y=1 \) gives \( z=1 \))
- Direction vector \( \vec{b_1} = (3, 1, 2) \) (coefficients of \( \lambda \))
For \( L_2 \):
- A point \( \vec{a_2} = (1, 0, 0) \) (substituting \( x=1, y=0 \) gives \( z=0 \))
- Direction vector \( \vec{b_2} = (1, 2, 3) \) (coefficients of \( \mu \))
### Step 3: Use the formula for shortest distance
The formula for the shortest distance \( d \) between two skew lines is given by:
\[
d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}
\]
### Step 4: Calculate \( \vec{a_2} - \vec{a_1} \)
\[
\vec{a_2} - \vec{a_1} = (1 - 1, 0 - 1, 0 - 1) = (0, -1, -1)
\]
### Step 5: Calculate \( \vec{b_1} \times \vec{b_2} \)
\[
\vec{b_1} = (3, 1, 2), \quad \vec{b_2} = (1, 2, 3)
\]
Using the determinant to find the cross product:
\[
\vec{b_1} \times \vec{b_2} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
3 & 1 & 2 \\
1 & 2 & 3
\end{vmatrix}
= \hat{i}(1 \cdot 3 - 2 \cdot 2) - \hat{j}(3 \cdot 3 - 2 \cdot 1) + \hat{k}(3 \cdot 2 - 1 \cdot 1)
\]
\[
= \hat{i}(3 - 4) - \hat{j}(9 - 2) + \hat{k}(6 - 1)
\]
\[
= -\hat{i} - 7\hat{j} + 5\hat{k} = (-1, -7, 5)
\]
### Step 6: Calculate the magnitude of \( \vec{b_1} \times \vec{b_2} \)
\[
|\vec{b_1} \times \vec{b_2}| = \sqrt{(-1)^2 + (-7)^2 + 5^2} = \sqrt{1 + 49 + 25} = \sqrt{75}
\]
### Step 7: Calculate the dot product
\[
(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (0, -1, -1) \cdot (-1, -7, 5)
\]
\[
= 0 \cdot -1 + (-1) \cdot -7 + (-1) \cdot 5 = 0 + 7 - 5 = 2
\]
### Step 8: Substitute into the distance formula
\[
d = \frac{|2|}{\sqrt{75}} = \frac{2}{\sqrt{75}} = \frac{2}{5\sqrt{3}} = \frac{2\sqrt{3}}{15}
\]
### Final Answer
The shortest distance between the two lines \( L_1 \) and \( L_2 \) is \( \frac{2\sqrt{3}}{15} \).
