A variable plane is at constant distance 3 from origin and meets the axis in `X,Y,Z` planes through `X,Y,Z` parallel to coordinate planes are made. If `P`be the point of intersection of above made planes, given `P(alpha, beta, gamma)` then `(alpha beta gamma)/(sqrt(alpha^(2)beta^(2)+beta^(2)gamma^(2)+gamma^(2)alpha^(2)))=?`

Explanation

To solve the problem step by step, we will derive the necessary equations and find the required value.

### Step 1: Understand the equation of the plane
The equation of a plane at a distance \( d \) from the origin can be expressed as:
\[ \frac{x}{A} + \frac{y}{B} + \frac{z}{C} = 1 \]
where \( A, B, C \) are the intercepts on the x, y, and z axes respectively.

### Step 2: Use the distance formula
The distance \( D \) from the origin (0, 0, 0) to the plane is given by:
\[ D = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
For our case, since the plane is at a constant distance of 3 from the origin, we have:
\[ 3 = \frac{|0 + 0 + 0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
This simplifies to:
\[ |D| = 3\sqrt{A^2 + B^2 + C^2} \]

### Step 3: Set up the plane equation
Assuming the plane equation is:
\[ \frac{x}{A} + \frac{y}{B} + \frac{z}{C} = 1 \]
We can rearrange it to:
\[ x + \frac{yB}{A} + \frac{zC}{A} - A = 0 \]
This implies \( D = -A \).

### Step 4: Substitute into the distance equation
Substituting \( D = -A \) into the distance equation gives:
\[ |-A| = 3\sqrt{A^2 + B^2 + C^2} \]
Thus:
\[ A = 3\sqrt{A^2 + B^2 + C^2} \]

### Step 5: Square both sides
Squaring both sides results in:
\[ A^2 = 9(A^2 + B^2 + C^2) \]
Rearranging gives:
\[ A^2(1 - 9) = 9(B^2 + C^2) \]
\[ -8A^2 = 9(B^2 + C^2) \]
This implies:
\[ \frac{1}{A^2} + \frac{1}{B^2} + \frac{1}{C^2} = \frac{1}{9} \]

### Step 6: Express in terms of \( \alpha, \beta, \gamma \)
Let \( \alpha = A, \beta = B, \gamma = C \). Then:
\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} = \frac{1}{9} \]

### Step 7: Find the required expression
We need to find:
\[ \frac{\alpha \beta \gamma}{\sqrt{\alpha^2 \beta^2 + \beta^2 \gamma^2 + \gamma^2 \alpha^2}} \]
From the previous result, we can express:
\[ \alpha^2 \beta^2 \gamma^2 = \frac{1}{\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2}} = 9 \]
Thus, \( \alpha \beta \gamma = 3 \).

### Step 8: Substitute back into the expression
Now we substitute back:
\[ \frac{3}{\sqrt{\alpha^2 \beta^2 + \beta^2 \gamma^2 + \gamma^2 \alpha^2}} = 3 \]

### Conclusion
Thus, the final answer is:
\[ \frac{\alpha \beta \gamma}{\sqrt{\alpha^2 \beta^2 + \beta^2 \gamma^2 + \gamma^2 \alpha^2}} = 3 \]