The equations to sphere passing through origin and the points `(-1,0,0),(0,-2,0) and (0,0,-3)` is `x^2+y^2+z^2+f(x,y,z)=0`. What is `f(x,y,z)=`

Options

A. A) `-x-2y-3z`
B. B) `x+2y+3z`
C. c) `x+2y+3z-1`
D. D) `x+2y+3z+1`

Answer

To find the function \( f(x, y, z) \) for the sphere passing through the origin and the points \((-1, 0, 0)\), \((0, -2, 0)\), and \((0, 0, -3)\), we can follow these steps:

### Step 1: Write the general equation of a sphere
The general equation of a sphere can be expressed as:
\[ x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0 \]
Since the sphere passes through the origin, we can substitute \( (0, 0, 0) \) into the equation.

### Step 2: Substitute the origin into the equation
Substituting \( (0, 0, 0) \):
\[ 0^2 + 0^2 + 0^2 + 2u(0) + 2v(0) + 2w(0) + d = 0 \]
This simplifies to:
\[ d = 0 \]

### Step 3: Substitute the point \((-1, 0, 0)\)
Next, we substitute the point \((-1, 0, 0)\):
\[ (-1)^2 + 0^2 + 0^2 + 2u(-1) + 2v(0) + 2w(0) + 0 = 0 \]
This simplifies to:
\[ 1 - 2u = 0 \implies u = \frac{1}{2} \]

### Step 4: Substitute the point \((0, -2, 0)\)
Now, we substitute the point \((0, -2, 0)\):
\[ 0^2 + (-2)^2 + 0^2 + 2u(0) + 2v(-2) + 2w(0) + 0 = 0 \]
This simplifies to:
\[ 4 - 4v = 0 \implies v = 1 \]

### Step 5: Substitute the point \((0, 0, -3)\)
Finally, we substitute the point \((0, 0, -3)\):
\[ 0^2 + 0^2 + (-3)^2 + 2u(0) + 2v(0) + 2w(-3) + 0 = 0 \]
This simplifies to:
\[ 9 - 6w = 0 \implies w = \frac{3}{2} \]

### Step 6: Write the function \( f(x, y, z) \)
Now that we have \( u \), \( v \), and \( w \), we can express \( f(x, y, z) \):
\[ f(x, y, z) = 2ux + 2vy + 2wz \]
Substituting the values:
\[ f(x, y, z) = 2\left(\frac{1}{2}\right)x + 2(1)y + 2\left(\frac{3}{2}\right)z \]
This simplifies to:
\[ f(x, y, z) = x + 2y + 3z \]

### Final Answer
The function \( f(x, y, z) \) is:
\[ f(x, y, z) = x + 2y + 3z \]