Three straight lines mutually perpendicular to each other meet in a point P and one of them intersects the x-axis and another intersects the y-axis, while the third line passes through a fixed point(0,0,c) on the z-axis. Then the locus of P is

Options

A. `x^(2)+y^(2)+z^(2)-2cx=0`
B. `x^(2)+y^(2)+z^(2)-2cy=0`
C. `x^(2)+y^(2)+z^(2)-2cz=0`
D. `x^(2)+y^(2)+z^(2)-2c(x+y+z)=0`

Explanation

To solve the problem, we need to find the locus of the point P where three mutually perpendicular lines intersect. One line intersects the x-axis, another intersects the y-axis, and the third line passes through a fixed point on the z-axis, specifically (0, 0, c).

### Step-by-Step Solution:

1. **Define the Points of Intersection:**
- Let the point where line L1 intersects the x-axis be A(a, 0, 0).
- Let the point where line L2 intersects the y-axis be B(0, b, 0).
- The line L3 intersects the z-axis at the fixed point C(0, 0, c).

2. **Coordinates of Point P:**
- Let P be the point of intersection of the three lines, which can be represented as P(x, y, z).

3. **Direction Ratios of the Lines:**
- The direction ratios (DRs) for line L1 can be represented as (x - a, y, z).
- The direction ratios for line L2 can be represented as (x, y - b, z).
- The direction ratios for line L3 can be represented as (x, y, z - c).

4. **Using the Condition of Perpendicularity:**
- Since the lines are mutually perpendicular, the dot product of the direction ratios of any two lines must equal zero.

5. **Equation from L1 and L2:**
- The dot product of L1 and L2:
\[ (x - a)(x) + (y)(y - b) + (z)(z) = 0 \]
- Expanding this gives:
\[ x^2 - ax + y^2 - by + z^2 = 0 \quad \text{(Equation 1)} \]

6. **Equation from L2 and L3:**
- The dot product of L2 and L3:
\[ (x)(x) + (y - b)(y) + (z)(z - c) = 0 \]
- Expanding this gives:
\[ x^2 + y^2 - by + z^2 - cz = 0 \quad \text{(Equation 2)} \]

7. **Equation from L1 and L3:**
- The dot product of L1 and L3:
\[ (x - a)(x) + (y)(y) + (z)(z - c) = 0 \]
- Expanding this gives:
\[ x^2 - ax + y^2 + z^2 - cz = 0 \quad \text{(Equation 3)} \]

8. **Combining Equations:**
- By adding Equation 2 and Equation 3, we can eliminate some terms:
\[ (x^2 + y^2 + z^2) - (ax + by + cz) = 0 \]
- Rearranging gives:
\[ x^2 + y^2 + z^2 - 2cz = 0 \]

9. **Final Locus Equation:**
- Replacing x, y, z with their respective coordinates gives us:
\[ x^2 + y^2 + z^2 - 2cz = 0 \]
- This represents the locus of point P.

### Conclusion:
The locus of point P is given by the equation:
\[ x^2 + y^2 + z^2 - 2cz = 0 \]