A plane passes through a fixed point `(a ,b ,c)dot` The locus of the foot of the perpendicular to it from the origin is a sphere of radius a. `1/2sqrt(a^2+b^2+c^2)` b. `sqrt(a^2+b^2+c^2)` c. `a^2+b^2+c^2` d. `1/2(a^2+b^2+c^2)`

Options

A. `(1)/(2)sqrt(a^(2)+b^(2)+c^(2))`
B. `sqrt(a^(2)+b^(2)+c^(2))`
C. `a^(2)+b^(2)+c^(2)`
D. `(1)/(2)(a^(2)+b^(2)+c^(2))`

Explanation

To solve the problem, we need to find the locus of the foot of the perpendicular from the origin to a plane that passes through the fixed point \((a, b, c)\).

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- We have a fixed point \(P(a, b, c)\) in space.
- We need to find the locus of the foot of the perpendicular from the origin \(O(0, 0, 0)\) to the plane that passes through point \(P\).

2. **Setting Up the Coordinates**:
- Let the foot of the perpendicular from the origin to the plane be denoted as \(F(\alpha, \beta, \gamma)\).

3. **Using the Condition for Perpendicularity**:
- The vector \( \overrightarrow{OP} = (a, b, c) \) and the vector \( \overrightarrow{OF} = (\alpha, \beta, \gamma) \).
- The line \(PF\) is perpendicular to the line \(OF\), which gives us the condition:
\[ \overrightarrow{OP} \cdot \overrightarrow{OF} = 0 \]
- This translates to:
\[ a(\alpha - 0) + b(\beta - 0) + c(\gamma - 0) = 0 \]
- Thus, we have:
\[ a\alpha + b\beta + c\gamma = 0 \]

4. **Equation of the Plane**:
- The equation of the plane can be expressed as:
\[ ax + by + cz = d \]
- Since the plane passes through the point \(P(a, b, c)\), we can find \(d\) by substituting \(x = a\), \(y = b\), and \(z = c\):
\[ d = aa + bb + cc = a^2 + b^2 + c^2 \]
- Therefore, the equation of the plane becomes:
\[ ax + by + cz = a^2 + b^2 + c^2 \]

5. **Finding the Locus**:
- The foot of the perpendicular \(F(\alpha, \beta, \gamma)\) must satisfy both the plane equation and the perpendicularity condition.
- From the perpendicularity condition:
\[ a\alpha + b\beta + c\gamma = 0 \]
- From the plane equation:
\[ a\alpha + b\beta + c\gamma = a^2 + b^2 + c^2 \]
- Setting these equal gives us a relationship that can be used to find the locus of points \(F\).

6. **Finding the Radius of the Sphere**:
- The locus of the foot of the perpendicular from the origin to the plane is a sphere.
- The center of this sphere can be found as the midpoint between the origin and the point \(P\), which is \(\left(\frac{a}{2}, \frac{b}{2}, \frac{c}{2}\right)\).
- The radius \(R\) of the sphere can be calculated as:
\[ R = \sqrt{\left(\frac{a}{2}\right)^2 + \left(\frac{b}{2}\right)^2 + \left(\frac{c}{2}\right)^2} = \frac{1}{2}\sqrt{a^2 + b^2 + c^2} \]

### Conclusion:
The radius of the sphere is \(\frac{1}{2}\sqrt{a^2 + b^2 + c^2}\).

### Final Answer:
The correct option is **(a)** \(\frac{1}{2}\sqrt{a^2 + b^2 + c^2}\).

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