To solve the problem, we need to find the distance from the point \( A(1, 2, 3) \) to the given line, measured parallel to the plane \( P \) whose normal line has positive direction cosines and makes equal angles with the coordinate axes.
### Step-by-step Solution:
1. **Identify the Direction Ratios of the Normal Line:**
Since the normal line makes equal angles with the coordinate axes, the direction ratios of the normal to the plane \( P \) can be taken as \( (1, 1, 1) \).
2. **Parameterize the Given Line:**
The line is given by the equation:
\[
\frac{x-1}{1} = \frac{y+2}{1} = \frac{z-3}{2} = \lambda
\]
From this, we can express the coordinates of any point \( B \) on the line in terms of \( \lambda \):
\[
B(\lambda + 1, \lambda - 2, 2\lambda + 3)
\]
3. **Find the Vector \( \overrightarrow{AB} \):**
The vector \( \overrightarrow{AB} \) from point \( A(1, 2, 3) \) to point \( B(\lambda + 1, \lambda - 2, 2\lambda + 3) \) is given by:
\[
\overrightarrow{AB} = B - A = (\lambda + 1 - 1, \lambda - 2 - 2, 2\lambda + 3 - 3) = (\lambda, \lambda - 4, 2\lambda)
\]
4. **Set Up the Perpendicular Condition:**
Since \( \overrightarrow{AB} \) is perpendicular to the normal line \( (1, 1, 1) \), we can use the dot product:
\[
\overrightarrow{AB} \cdot (1, 1, 1) = 0
\]
This gives us:
\[
\lambda + (\lambda - 4) + 2\lambda = 0
\]
Simplifying this, we get:
\[
4\lambda - 4 = 0 \implies 4\lambda = 4 \implies \lambda = 1
\]
5. **Find the Coordinates of Point \( B \):**
Substitute \( \lambda = 1 \) back into the parameterization of the line:
\[
B(1 + 1, 1 - 2, 2 \cdot 1 + 3) = (2, -1, 5)
\]
6. **Calculate the Vector \( \overrightarrow{AB} \) Again:**
Now, we can find \( \overrightarrow{AB} \):
\[
\overrightarrow{AB} = (2 - 1, -1 - 2, 5 - 3) = (1, -3, 2)
\]
7. **Find the Magnitude of \( \overrightarrow{AB} \):**
The distance \( d \) from point \( A \) to the line, measured parallel to the plane \( P \), is the magnitude of \( \overrightarrow{AB} \):
\[
d = |\overrightarrow{AB}| = \sqrt{1^2 + (-3)^2 + 2^2} = \sqrt{1 + 9 + 4} = \sqrt{14}
\]
### Final Answer:
The distance of the point \( A(1, 2, 3) \) from the line, measured parallel to the plane \( P \), is \( \sqrt{14} \).
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