A point `P` moves on a plane `x/a+y/b+z/c=1.` A plane through `P` and perpendicular to `O P` meets the coordinate axes at `A , Ba n d Cdot` If the planes through `A ,Ba n dC` parallel to the planes `x=0,y=0a n dz=0,` respectively, intersect at `Q ,` find the locus of `Qdot`

Explanation

To solve the problem, we need to find the locus of the point \( Q \) given the conditions in the question. Let's break down the solution step by step.

### Step 1: Write the equation of the plane
The plane is given by the equation:
\[ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \]

### Step 2: Assume the coordinates of point \( P \)
Let the coordinates of point \( P \) be \( (H, K, L) \). Since point \( P \) lies on the plane, it satisfies the equation:
\[ \frac{H}{a} + \frac{K}{b} + \frac{L}{c} = 1 \tag{1} \]

### Step 3: Find the distance from the origin to point \( P \)
The distance \( OP \) from the origin \( O(0, 0, 0) \) to point \( P(H, K, L) \) is given by:
\[ OP = \sqrt{H^2 + K^2 + L^2} \]

### Step 4: Find the direction cosines of \( OP \)
The direction cosines \( l, m, n \) of the line \( OP \) are given by:
\[ l = \frac{H}{\sqrt{H^2 + K^2 + L^2}}, \quad m = \frac{K}{\sqrt{H^2 + K^2 + L^2}}, \quad n = \frac{L}{\sqrt{H^2 + K^2 + L^2}} \]

### Step 5: Write the equation of the plane through \( P \) and perpendicular to \( OP \)
The equation of the plane passing through point \( P(H, K, L) \) and perpendicular to \( OP \) is:
\[ \frac{H}{\sqrt{H^2 + K^2 + L^2}}(x - H) + \frac{K}{\sqrt{H^2 + K^2 + L^2}}(y - K) + \frac{L}{\sqrt{H^2 + K^2 + L^2}}(z - L) = 0 \]
This simplifies to:
\[ Hx + Ky + Lz = H^2 + K^2 + L^2 \tag{2} \]

### Step 6: Find the coordinates of points \( A, B, C \)
The plane intersects the coordinate axes at points \( A, B, C \):
- Point \( A \) (on x-axis): Set \( y = 0 \) and \( z = 0 \):
\[ A = \left( \frac{H^2 + K^2 + L^2}{H}, 0, 0 \right) \]
- Point \( B \) (on y-axis): Set \( x = 0 \) and \( z = 0 \):
\[ B = \left( 0, \frac{H^2 + K^2 + L^2}{K}, 0 \right) \]
- Point \( C \) (on z-axis): Set \( x = 0 \) and \( y = 0 \):
\[ C = \left( 0, 0, \frac{H^2 + K^2 + L^2}{L} \right) \]

### Step 7: Define the coordinates of point \( Q \)
Let the coordinates of point \( Q \) be \( (x, y, z) \). The coordinates of \( Q \) can be expressed in terms of the coordinates of points \( A, B, C \):
\[ x = \frac{H^2 + K^2 + L^2}{H}, \quad y = \frac{H^2 + K^2 + L^2}{K}, \quad z = \frac{H^2 + K^2 + L^2}{L} \]

### Step 8: Express \( H, K, L \) in terms of \( x, y, z \)
From the expressions for \( x, y, z \), we can write:
\[ H = \frac{H^2 + K^2 + L^2}{x}, \quad K = \frac{H^2 + K^2 + L^2}{y}, \quad L = \frac{H^2 + K^2 + L^2}{z} \]
Let \( D = H^2 + K^2 + L^2 \). Then:
\[ H = \frac{D}{x}, \quad K = \frac{D}{y}, \quad L = \frac{D}{z} \]

### Step 9: Substitute back into the plane equation
Substituting \( H, K, L \) back into equation (1):
\[ \frac{D/x}{a} + \frac{D/y}{b} + \frac{D/z}{c} = 1 \]
This simplifies to:
\[ D \left( \frac{1}{ax} + \frac{1}{by} + \frac{1}{cz} \right) = 1 \]

### Step 10: Relate \( D \) to \( x, y, z \)
Since \( D = H^2 + K^2 + L^2 \), we can express it as:
\[ D = \frac{D^2}{x^2} + \frac{D^2}{y^2} + \frac{D^2}{z^2} \]
Thus, we have:
\[ \frac{1}{ax} + \frac{1}{by} + \frac{1}{cz} = \frac{1}{D} \]

### Final Step: Locus of \( Q \)
The locus of point \( Q \) is given by:
\[ \frac{1}{ax} + \frac{1}{by} + \frac{1}{cz} = \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} \]