If the foot of the perpendicular from the origin to plane is `P(a ,b ,c)` , the equation of the plane is a. `x/a=y/b=z/c=3` b. `a x+b y+c z=3` c. `a x+b y+c z=a^(2)+b^2+c^2` d. `a x+b y+c z=a+b+c`

Options

A. `(x)/(a)+(y)/(b)+(z)/(c)=3`
B. `ax+by+cz=3`
C. `ax+by+cz=a^(2)+b^(2)+c^(2)`
D. `ax+by+cz=a+b+c`

Explanation

To solve the question, we need to find the equation of a plane given that the foot of the perpendicular from the origin to the plane is the point \( P(a, b, c) \).

### Step-by-step Solution:

1. **Understanding the Problem**:
The foot of the perpendicular from the origin (0, 0, 0) to the plane is given as the point \( P(a, b, c) \). This means that the vector from the origin to point \( P \) is normal to the plane.

2. **Normal Vector**:
The normal vector \( \mathbf{n} \) to the plane can be represented as:
\[ \mathbf{n} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \]
where \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) are the unit vectors along the x, y, and z axes respectively.

3. **General Equation of the Plane**:
The general equation of a plane can be expressed in the form:
\[ ax + by + cz = d \]
where \( (a, b, c) \) are the coefficients corresponding to the normal vector.

4. **Finding the Value of \( d \)**:
Since the point \( P(a, b, c) \) lies on the plane, we can substitute these coordinates into the plane equation:
\[ a(a) + b(b) + c(c) = d \]
This simplifies to:
\[ d = a^2 + b^2 + c^2 \]

5. **Final Equation of the Plane**:
Substituting \( d \) back into the plane equation gives:
\[ ax + by + cz = a^2 + b^2 + c^2 \]
Therefore, the equation of the plane is:
\[ ax + by + cz = a^2 + b^2 + c^2 \]

6. **Identifying the Correct Option**:
Among the given options, the correct equation of the plane is:
\[ ax + by + cz = a^2 + b^2 + c^2 \]
Thus, the answer is option **C**.