To solve the problem, we need to find the value of \( \alpha \) given that the foot of the perpendicular from the point \( (2, 5, 1) \) to the line passing through the point \( (\alpha, 2\alpha, 5) \) is \( \left(\frac{1}{5}, \frac{2}{5}, \frac{3}{5}\right) \).
### Step 1: Define the points
Let:
- Point A (the given point) = \( (2, 5, 1) \)
- Point B (the foot of the perpendicular) = \( \left(\frac{1}{5}, \frac{2}{5}, \frac{3}{5}\right) \)
- Point C (a point on the line) = \( (\alpha, 2\alpha, 5) \)
### Step 2: Find the vector AB
The vector \( \overrightarrow{AB} \) is given by:
\[
\overrightarrow{AB} = B - A = \left(\frac{1}{5} - 2, \frac{2}{5} - 5, \frac{3}{5} - 1\right)
\]
Calculating each component:
\[
\overrightarrow{AB} = \left(\frac{1}{5} - \frac{10}{5}, \frac{2}{5} - \frac{25}{5}, \frac{3}{5} - \frac{5}{5}\right) = \left(-\frac{9}{5}, -\frac{23}{5}, -\frac{2}{5}\right)
\]
### Step 3: Find the vector BC
The vector \( \overrightarrow{BC} \) is given by:
\[
\overrightarrow{BC} = C - B = \left(\alpha - \frac{1}{5}, 2\alpha - \frac{2}{5}, 5 - \frac{3}{5}\right)
\]
Calculating each component:
\[
\overrightarrow{BC} = \left(\alpha - \frac{1}{5}, 2\alpha - \frac{2}{5}, \frac{25}{5} - \frac{3}{5}\right) = \left(\alpha - \frac{1}{5}, 2\alpha - \frac{2}{5}, \frac{22}{5}\right)
\]
### Step 4: Use the dot product
Since \( \overrightarrow{AB} \) and \( \overrightarrow{BC} \) are perpendicular, their dot product is zero:
\[
\overrightarrow{AB} \cdot \overrightarrow{BC} = 0
\]
Calculating the dot product:
\[
\left(-\frac{9}{5}, -\frac{23}{5}, -\frac{2}{5}\right) \cdot \left(\alpha - \frac{1}{5}, 2\alpha - \frac{2}{5}, \frac{22}{5}\right) = 0
\]
Expanding this:
\[
-\frac{9}{5}\left(\alpha - \frac{1}{5}\right) - \frac{23}{5}\left(2\alpha - \frac{2}{5}\right) - \frac{2}{5}\left(\frac{22}{5}\right) = 0
\]
This simplifies to:
\[
-\frac{9\alpha}{5} + \frac{9}{25} - \frac{46\alpha}{5} + \frac{46}{25} - \frac{44}{25} = 0
\]
Combining like terms:
\[
-\frac{55\alpha}{5} + \frac{9 + 46 - 44}{25} = 0
\]
\[
-11\alpha + \frac{11}{25} = 0
\]
### Step 5: Solve for \( \alpha \)
Rearranging gives:
\[
11\alpha = \frac{11}{25}
\]
Dividing both sides by 11:
\[
\alpha = \frac{1}{25}
\]
### Final Answer
Thus, the value of \( \alpha \) is:
\[
\alpha = \frac{1}{25}
\]
