Let `pi_(1)and pi_(2)` be two planes and `L` be a line such that
`pi_(1): x + 2y + 3z = 14`
`pi_(2) : 2x - y +3z = 27`
`L : (x+1)/2=(y+1)/3=(z+1)/4`
If the line `L` meets the plane `pi_(1)` in `P(alpha, beta, gamma),`
Then `alpha^(2)+beta^(2)+gamma^(2)` is equal to:

Options

A. (a) `2`
B. (b) `14`
C. (c) `28`
D. (d) `29`

Answer

To solve the problem, we need to find the point \( P(\alpha, \beta, \gamma) \) where the line \( L \) intersects the plane \( \pi_1 \), and then calculate \( \alpha^2 + \beta^2 + \gamma^2 \).

### Step-by-Step Solution:

1. **Define the Line and Planes**:
- The line \( L \) is given by:
\[ \frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z + 1}{4} \]
- The plane \( \pi_1 \) is defined by:
\[ x + 2y + 3z = 14 \]

2. **Parameterize the Line**:
- Let \( \frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z + 1}{4} = \lambda \) (where \( \lambda \) is a parameter).
- From this, we can express \( x, y, z \) in terms of \( \lambda \):
\[ x = 2\lambda - 1, \quad y = 3\lambda - 1, \quad z = 4\lambda - 1 \]

3. **Substitute into the Plane Equation**:
- Substitute \( x, y, z \) into the plane equation \( \pi_1 \):
\[ (2\lambda - 1) + 2(3\lambda - 1) + 3(4\lambda - 1) = 14 \]
- Simplifying this:
\[ 2\lambda - 1 + 6\lambda - 2 + 12\lambda - 3 = 14 \]
\[ 20\lambda - 6 = 14 \]
\[ 20\lambda = 20 \implies \lambda = 1 \]

4. **Find Coordinates \( \alpha, \beta, \gamma \)**:
- Substitute \( \lambda = 1 \) back into the parameterized equations:
\[ \alpha = 2(1) - 1 = 1 \]
\[ \beta = 3(1) - 1 = 2 \]
\[ \gamma = 4(1) - 1 = 3 \]

5. **Calculate \( \alpha^2 + \beta^2 + \gamma^2 \)**:
- Now compute:
\[ \alpha^2 + \beta^2 + \gamma^2 = 1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14 \]

### Conclusion:
The value of \( \alpha^2 + \beta^2 + \gamma^2 \) is \( 14 \).