Let P be a plane passing through the points (a,0,0),(0,b,0),(0,0,c) then the image of origin with respect to plane P is :

Options

A. `[2/(a(1/a^2+1/b^2+1/c^2)),2/(b(1/a^2+1/b^2+1/c^2)),2/(c(1/a^2+1/b^2+1/c^2))]`
B. `(1/(1/a^2+1/b^2+1/c^2),1/(1/a^2+1/b^2+1/c^2),1/(1/a^2+1/b^2+1/c^2))`
C. (-2a,-2b,-2c)
D. None of these

Explanation

To find the image of the origin with respect to the plane \( P \) that passes through the points \( (a, 0, 0) \), \( (0, b, 0) \), and \( (0, 0, c) \), we can follow these steps:

### Step 1: Determine the equation of the plane
The plane \( P \) can be expressed in intercept form as:
\[ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \]
This can be rearranged to:
\[ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} - 1 = 0 \]

### Step 2: Identify coefficients for the image formula
From the equation of the plane, we can identify the coefficients:
- \( A = \frac{1}{a} \)
- \( B = \frac{1}{b} \)
- \( C = \frac{1}{c} \)
- \( D = 1 \)

### Step 3: Use the image formula
The formula for the image of a point \( (x_1, y_1, z_1) \) with respect to the plane \( Ax + By + Cz + D = 0 \) is given by:
\[ \frac{x - x_1}{A} = \frac{y - y_1}{B} = \frac{z - z_1}{C} = -\frac{2(Ax_1 + By_1 + Cz_1 + D)}{A^2 + B^2 + C^2} \]
For the origin \( (0, 0, 0) \):
- \( x_1 = 0 \)
- \( y_1 = 0 \)
- \( z_1 = 0 \)

### Step 4: Substitute values into the formula
Substituting into the formula:
\[ \frac{x - 0}{\frac{1}{a}} = \frac{y - 0}{\frac{1}{b}} = \frac{z - 0}{\frac{1}{c}} = -\frac{2\left(0 + 0 + 0 + 1\right)}{\left(\frac{1}{a}\right)^2 + \left(\frac{1}{b}\right)^2 + \left(\frac{1}{c}\right)^2} \]
This simplifies to:
\[ \frac{x}{\frac{1}{a}} = \frac{y}{\frac{1}{b}} = \frac{z}{\frac{1}{c}} = -\frac{2}{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}} \]

### Step 5: Solve for the coordinates of the image
Let \( k = -\frac{2}{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}} \). Then:
\[ x = k \cdot \frac{1}{a} = -\frac{2}{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}} \cdot \frac{1}{a} \]
\[ y = k \cdot \frac{1}{b} = -\frac{2}{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}} \cdot \frac{1}{b} \]
\[ z = k \cdot \frac{1}{c} = -\frac{2}{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}} \cdot \frac{1}{c} \]

### Final Result
Thus, the coordinates of the image of the origin with respect to the plane \( P \) are:
\[ \left(-\frac{2/a}{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}, -\frac{2/b}{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}, -\frac{2/c}{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}\right) \]