To find the direction cosines of the normal to the plane containing the lines \( L_1 \) and \( L_2 \), we can follow these steps:
### Step 1: Identify the direction ratios of the lines
The first line \( L_1 \) is given by:
\[
\frac{x-1}{2} = \frac{y-1}{1} = z-2
\]
From this, we can extract the direction ratios of \( L_1 \) as \( (2, 1, 0) \).
The second line \( L_2 \) is given by:
\[
\frac{x-2}{1} = \frac{y-3}{2} = \frac{z-2}{3}
\]
From this, we can extract the direction ratios of \( L_2 \) as \( (1, 2, 3) \).
### Step 2: Represent the direction ratios as vectors
Let:
\[
\mathbf{b_1} = (2, 1, 0) \quad \text{and} \quad \mathbf{b_2} = (1, 2, 3)
\]
### Step 3: Find the normal vector to the plane using the cross product
The normal vector \( \mathbf{n} \) can be found using the cross product of \( \mathbf{b_1} \) and \( \mathbf{b_2} \):
\[
\mathbf{n} = \mathbf{b_1} \times \mathbf{b_2}
\]
Calculating the cross product:
\[
\mathbf{n} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 1 & 0 \\
1 & 2 & 3
\end{vmatrix}
\]
Calculating the determinant:
\[
\mathbf{n} = \mathbf{i} \begin{vmatrix} 1 & 0 \\ 2 & 3 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 2 & 0 \\ 1 & 3 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix}
\]
\[
= \mathbf{i} (1 \cdot 3 - 0 \cdot 2) - \mathbf{j} (2 \cdot 3 - 0 \cdot 1) + \mathbf{k} (2 \cdot 2 - 1 \cdot 1)
\]
\[
= 3\mathbf{i} - 6\mathbf{j} + 3\mathbf{k}
\]
Thus, the normal vector is:
\[
\mathbf{n} = (3, -6, 3)
\]
### Step 4: Calculate the magnitude of the normal vector
The magnitude \( |\mathbf{n}| \) is given by:
\[
|\mathbf{n}| = \sqrt{3^2 + (-6)^2 + 3^2} = \sqrt{9 + 36 + 9} = \sqrt{54} = 3\sqrt{6}
\]
### Step 5: Find the direction cosines
The direction cosines \( (\cos \alpha, \cos \beta, \cos \gamma) \) are given by:
\[
\cos \alpha = \frac{n_1}{|\mathbf{n}|}, \quad \cos \beta = \frac{n_2}{|\mathbf{n}|}, \quad \cos \gamma = \frac{n_3}{|\mathbf{n}|}
\]
Substituting the values:
\[
\cos \alpha = \frac{3}{3\sqrt{6}} = \frac{1}{\sqrt{6}}, \quad \cos \beta = \frac{-6}{3\sqrt{6}} = -\frac{2}{\sqrt{6}}, \quad \cos \gamma = \frac{3}{3\sqrt{6}} = \frac{1}{\sqrt{6}}
\]
Thus, the direction cosines of the normal to the plane are:
\[
\left( \frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}} \right)
\]
### Final Answer:
The direction cosines of the normal to the plane are:
\[
\left( \frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}} \right)
\]
