Let the image of the point P(1,2,3) in the line `L : (x-6)/(3)= (y-1)/(2)= (z-2)/(3)` be Q. Let `R (alpha, beta, gamma)` be a point that divides internally the line segment PQ in the ratio 1:3. Then the value of `22 (alpha + beta + gamma)` is equal to _____

Answer

To solve the problem, we need to find the coordinates of the point Q, which is the image of point P(1, 2, 3) with respect to the line L given by the equation \((x-6)/3 = (y-1)/2 = (z-2)/3\). Then, we will find the coordinates of point R that divides the segment PQ in the ratio 1:3, and finally compute \(22(\alpha + \beta + \gamma)\).

### Step-by-step Solution:

1. **Identify the direction ratios of the line L:**
The line L can be expressed in parametric form as:
\[ x = 6 + 3t, \quad y = 1 + 2t, \quad z = 2 + 3t \]
The direction ratios of the line are (3, 2, 3).

2. **Find the coordinates of the foot of the perpendicular from P to the line L:**
Let the coordinates of the foot of the perpendicular be \(T(x_T, y_T, z_T)\). The coordinates can be expressed as:
\[ x_T = 6 + 3\lambda, \quad y_T = 1 + 2\lambda, \quad z_T = 2 + 3\lambda \]
where \(\lambda\) is a parameter.

3. **Calculate the direction ratios of PT:**
The direction ratios of the line segment PT are:
\[ (x_T - 1, y_T - 2, z_T - 3) = (3\lambda + 5, 2\lambda - 1, 3\lambda - 1) \]

4. **Set up the dot product condition for perpendicularity:**
The line PT is perpendicular to the line L, so their dot product must equal zero:
\[ (3\lambda + 5) \cdot 3 + (2\lambda - 1) \cdot 2 + (3\lambda - 1) \cdot 3 = 0 \]
Expanding this gives:
\[ 9\lambda + 15 + 4\lambda - 2 + 9\lambda - 3 = 0 \]
Simplifying:
\[ 22\lambda + 10 = 0 \implies \lambda = -\frac{10}{22} = -\frac{5}{11} \]

5. **Substitute \(\lambda\) back to find coordinates of T:**
\[ x_T = 6 + 3\left(-\frac{5}{11}\right) = 6 - \frac{15}{11} = \frac{66 - 15}{11} = \frac{51}{11} \]
\[ y_T = 1 + 2\left(-\frac{5}{11}\right) = 1 - \frac{10}{11} = \frac{1 - 10}{11} = -\frac{9}{11} \]
\[ z_T = 2 + 3\left(-\frac{5}{11}\right) = 2 - \frac{15}{11} = \frac{22 - 15}{11} = \frac{7}{11} \]
Thus, the coordinates of T are \(\left(\frac{51}{11}, -\frac{9}{11}, \frac{7}{11}\right)\).

6. **Find the coordinates of Q:**
Since T is the midpoint of PQ, we can express the coordinates of Q as:
\[ \left(\frac{x_Q + 1}{2}, \frac{y_Q + 2}{2}, \frac{z_Q + 3}{2}\right) = \left(\frac{51}{11}, -\frac{9}{11}, \frac{7}{11}\right) \]
Solving for \(x_Q, y_Q, z_Q\):
\[ x_Q + 1 = \frac{102}{11} \implies x_Q = \frac{102}{11} - 1 = \frac{102 - 11}{11} = \frac{91}{11} \]
\[ y_Q + 2 = -\frac{18}{11} \implies y_Q = -\frac{18}{11} - 2 = -\frac{18 + 22}{11} = -\frac{40}{11} \]
\[ z_Q + 3 = \frac{14}{11} \implies z_Q = \frac{14}{11} - 3 = \frac{14 - 33}{11} = -\frac{19}{11} \]
Thus, the coordinates of Q are \(\left(\frac{91}{11}, -\frac{40}{11}, -\frac{19}{11}\right)\).

7. **Find the coordinates of R:**
R divides PQ in the ratio 1:3, so using the section formula:
\[ \alpha = \frac{1 \cdot \frac{91}{11} + 3 \cdot 1}{1 + 3} = \frac{\frac{91}{11} + 3}{4} = \frac{\frac{91 + 33}{11}}{4} = \frac{124/11}{4} = \frac{31}{11} \]
\[ \beta = \frac{1 \cdot \left(-\frac{40}{11}\right) + 3 \cdot 2}{4} = \frac{-\frac{40}{11} + 6}{4} = \frac{-\frac{40}{11} + \frac{66}{11}}{4} = \frac{26/11}{4} = \frac{13}{22} \]
\[ \gamma = \frac{1 \cdot \left(-\frac{19}{11}\right) + 3 \cdot 3}{4} = \frac{-\frac{19}{11} + 9}{4} = \frac{-\frac{19}{11} + \frac{99}{11}}{4} = \frac{80/11}{4} = \frac{20}{11} \]

8. **Calculate \(22(\alpha + \beta + \gamma)\):**
\[ \alpha + \beta + \gamma = \frac{31}{11} + \frac{13}{22} + \frac{20}{11} \]
To add these, we need a common denominator (which is 22):
\[ \alpha + \beta + \gamma = \frac{62}{22} + \frac{13}{22} + \frac{40}{22} = \frac{115}{22} \]
Therefore,
\[ 22(\alpha + \beta + \gamma) = 22 \cdot \frac{115}{22} = 115 \]

### Final Answer:
The value of \(22(\alpha + \beta + \gamma)\) is **115**.