To solve the problem, we need to find the value of \( \alpha \) such that the point \( Q(5/3, 7/3, 17/3) \) is the foot of the perpendicular from the point \( P(1, 0, 3) \) to the line \( L \) that passes through the point \( R(\alpha, 7, 1) \).
### Step-by-Step Solution:
1. **Determine the vectors:**
- The vector \( \overrightarrow{QP} \) from \( Q \) to \( P \) is given by:
\[
\overrightarrow{QP} = P - Q = \left(1 - \frac{5}{3}, 0 - \frac{7}{3}, 3 - \frac{17}{3}\right)
\]
- Simplifying this:
\[
\overrightarrow{QP} = \left(1 - \frac{5}{3}, 0 - \frac{7}{3}, 3 - \frac{17}{3}\right) = \left(-\frac{2}{3}, -\frac{7}{3}, -\frac{8}{3}\right)
\]
2. **Determine the vector \( \overrightarrow{QR} \):**
- The vector \( \overrightarrow{QR} \) from \( Q \) to \( R \) is given by:
\[
\overrightarrow{QR} = R - Q = \left(\alpha - \frac{5}{3}, 7 - \frac{7}{3}, 1 - \frac{17}{3}\right)
\]
- Simplifying this:
\[
\overrightarrow{QR} = \left(\alpha - \frac{5}{3}, \frac{21}{3} - \frac{7}{3}, \frac{3}{3} - \frac{17}{3}\right) = \left(\alpha - \frac{5}{3}, \frac{14}{3}, -\frac{14}{3}\right)
\]
3. **Set up the dot product:**
- Since \( \overrightarrow{QP} \) and \( \overrightarrow{QR} \) are perpendicular, their dot product must equal zero:
\[
\overrightarrow{QP} \cdot \overrightarrow{QR} = 0
\]
- Calculating the dot product:
\[
\left(-\frac{2}{3}\right)\left(\alpha - \frac{5}{3}\right) + \left(-\frac{7}{3}\right)\left(\frac{14}{3}\right) + \left(-\frac{8}{3}\right)\left(-\frac{14}{3}\right) = 0
\]
4. **Expand and simplify the equation:**
- Expanding the dot product:
\[
-\frac{2}{3}(\alpha - \frac{5}{3}) - \frac{98}{9} + \frac{112}{9} = 0
\]
- This simplifies to:
\[
-\frac{2}{3}(\alpha - \frac{5}{3}) + \frac{14}{9} = 0
\]
5. **Isolate \( \alpha \):**
- Rearranging gives:
\[
-\frac{2}{3}(\alpha - \frac{5}{3}) = -\frac{14}{9}
\]
- Multiplying both sides by \(-\frac{3}{2}\):
\[
\alpha - \frac{5}{3} = \frac{21}{9}
\]
- Converting \( \frac{21}{9} \) to a simpler form:
\[
\frac{21}{9} = \frac{7}{3}
\]
- Thus:
\[
\alpha - \frac{5}{3} = \frac{7}{3}
\]
- Adding \( \frac{5}{3} \) to both sides:
\[
\alpha = \frac{7}{3} + \frac{5}{3} = \frac{12}{3} = 4
\]
### Final Answer:
\[
\alpha = 4
\]
