If the sum of squares of distances of a point from the planes `x+y+z= 0, x-z= 0 and x-2y +z= 0` is `p^(2)`, then locus of the point is

Options

A. `x^(2) +z^(2)=p^(2)`
B. `x^(2)+2xy + y^(2) + z^(2)=p^(2)`
C. `x+y+ z= p^(2)`
D. `x^(2) + y^(2) + z^(2) = p^(2)`

Answer

To solve the problem, we need to find the locus of a point \( P(\alpha, \beta, \gamma) \) such that the sum of the squares of the distances from the point to the given planes is equal to \( p^2 \).

### Step 1: Determine the distances from the point to each plane

1. **Distance from the plane \( x + y + z = 0 \)**:
The formula for the distance from a point \( (x_0, y_0, z_0) \) to the plane \( Ax + By + Cz + D = 0 \) is given by:
\[ \text{Distance} = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
For the plane \( x + y + z = 0 \), we have \( A = 1, B = 1, C = 1, D = 0 \). Thus, the distance is:
\[ d_1 = \frac{|\alpha + \beta + \gamma|}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{|\alpha + \beta + \gamma|}{\sqrt{3}} \]

2. **Distance from the plane \( x - z = 0 \)**:
Here, \( A = 1, B = 0, C = -1, D = 0 \). The distance is:
\[ d_2 = \frac{|\alpha - \gamma|}{\sqrt{1^2 + 0^2 + (-1)^2}} = \frac{|\alpha - \gamma|}{\sqrt{2}} \]

3. **Distance from the plane \( x - 2y + z = 0 \)**:
For this plane, \( A = 1, B = -2, C = 1, D = 0 \). The distance is:
\[ d_3 = \frac{|\alpha - 2\beta + \gamma|}{\sqrt{1^2 + (-2)^2 + 1^2}} = \frac{|\alpha - 2\beta + \gamma|}{\sqrt{6}} \]

### Step 2: Sum of the squares of the distances

The problem states that the sum of the squares of these distances is equal to \( p^2 \):
\[ d_1^2 + d_2^2 + d_3^2 = p^2 \]

Substituting the distances:
\[ \left(\frac{|\alpha + \beta + \gamma|}{\sqrt{3}}\right)^2 + \left(\frac{|\alpha - \gamma|}{\sqrt{2}}\right)^2 + \left(\frac{|\alpha - 2\beta + \gamma|}{\sqrt{6}}\right)^2 = p^2 \]

This simplifies to:
\[ \frac{(\alpha + \beta + \gamma)^2}{3} + \frac{(\alpha - \gamma)^2}{2} + \frac{(\alpha - 2\beta + \gamma)^2}{6} = p^2 \]

### Step 3: Clear the denominators

To eliminate the fractions, multiply through by the least common multiple of the denominators, which is 6:
\[ 2(\alpha + \beta + \gamma)^2 + 3(\alpha - \gamma)^2 + (\alpha - 2\beta + \gamma)^2 = 6p^2 \]

### Step 4: Expand and simplify

Now we expand each term:
1. \( 2(\alpha + \beta + \gamma)^2 = 2(\alpha^2 + \beta^2 + \gamma^2 + 2\alpha\beta + 2\alpha\gamma + 2\beta\gamma) \)
2. \( 3(\alpha - \gamma)^2 = 3(\alpha^2 + \gamma^2 - 2\alpha\gamma) \)
3. \( (\alpha - 2\beta + \gamma)^2 = \alpha^2 + 4\beta^2 + \gamma^2 - 4\alpha\beta + 2\alpha\gamma - 4\beta\gamma \)

Combining all these gives:
\[ (2 + 3 + 1)\alpha^2 + (2 + 4)\beta^2 + (2 + 3)\gamma^2 + (2 + 2 - 4)\alpha\beta + (2 - 2)\alpha\gamma + (-4)\beta\gamma = 6p^2 \]
This simplifies to:
\[ 6\alpha^2 + 6\beta^2 + 6\gamma^2 = 6p^2 \]

### Step 5: Final equation

Dividing through by 6, we get:
\[ \alpha^2 + \beta^2 + \gamma^2 = p^2 \]

### Conclusion

Thus, the locus of the point \( P(\alpha, \beta, \gamma) \) is a sphere centered at the origin with radius \( p \).