The plane passing through the point (-2, -2, -3)
and containing the line joining the points (1, 1, 1)
and (1, -1, 2) makes an intercept a, b, c along x,
y and z axis respectively, then `1/a+1/b+1/c` will be

Options

A. 3
B. 4
C. 6
D. 1

Answer

To solve the problem, we need to find the equation of the plane that passes through the point (-2, -2, -3) and contains the line joining the points (1, 1, 1) and (1, -1, 2). Then we will find the intercepts on the x, y, and z axes, and finally compute \( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \).

### Step-by-Step Solution:

1. **Identify the Points**:
- Let \( P = (-2, -2, -3) \) be the point through which the plane passes.
- Let \( A = (1, 1, 1) \) and \( B = (1, -1, 2) \) be the points that define the line.

2. **Find Direction Vector of Line AB**:
- The direction vector \( \vec{AB} \) can be calculated as:
\[ \vec{AB} = B - A = (1 - 1, -1 - 1, 2 - 1) = (0, -2, 1) \]

3. **Find a Normal Vector to the Plane**:
- We need another vector in the plane. We can use the vector from point \( P \) to point \( A \):
\[ \vec{PA} = A - P = (1 - (-2), 1 - (-2), 1 - (-3)) = (3, 3, 4) \]
- Now we have two vectors in the plane: \( \vec{AB} = (0, -2, 1) \) and \( \vec{PA} = (3, 3, 4) \).
- The normal vector \( \vec{n} \) to the plane can be found using the cross product:
\[ \vec{n} = \vec{AB} \times \vec{PA} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -2 & 1 \\ 3 & 3 & 4 \end{vmatrix} \]
- Calculating the determinant:
\[ \vec{n} = \hat{i}((-2)(4) - (1)(3)) - \hat{j}((0)(4) - (1)(3)) + \hat{k}((0)(3) - (-2)(3)) \]
\[ = \hat{i}(-8 - 3) - \hat{j}(0 - 3) + \hat{k}(0 + 6) \]
\[ = \hat{i}(-11) + \hat{j}(3) + \hat{k}(6) = (-11, 3, 6) \]

4. **Equation of the Plane**:
- The equation of the plane can be expressed as:
\[ -11(x + 2) + 3(y + 2) + 6(z + 3) = 0 \]
- Expanding this:
\[ -11x - 22 + 3y + 6 + 6z + 18 = 0 \]
\[ -11x + 3y + 6z - 22 + 24 = 0 \]
\[ -11x + 3y + 6z + 2 = 0 \]
- Rearranging gives:
\[ 11x - 3y - 6z = 2 \]

5. **Find Intercepts**:
- **X-intercept (a)**: Set \( y = 0 \) and \( z = 0 \):
\[ 11x = 2 \implies x = \frac{2}{11} \implies a = \frac{2}{11} \]
- **Y-intercept (b)**: Set \( x = 0 \) and \( z = 0 \):
\[ -3y = 2 \implies y = -\frac{2}{3} \implies b = -\frac{2}{3} \]
- **Z-intercept (c)**: Set \( x = 0 \) and \( y = 0 \):
\[ 6z = 2 \implies z = \frac{1}{3} \implies c = \frac{1}{3} \]

6. **Calculate \( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \)**:
- Substitute the values of \( a, b, c \):
\[ \frac{1}{a} = \frac{11}{2}, \quad \frac{1}{b} = -\frac{3}{2}, \quad \frac{1}{c} = 3 \]
- Now, calculate:
\[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{11}{2} - \frac{3}{2} + 3 \]
\[ = \frac{11 - 3 + 6}{2} = \frac{14}{2} = 7 \]

### Final Answer:
The value of \( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \) is \( 7 \).