The points in which the line
`(x-1)/(1)=(y-1)/(-1)=(z+3)/(1)` cuts the surface `x^(2)+y^(2)+z^(2)-20=0`

Options

A. (0, 2, 4)
B. (0, 1, -4)
C. (4, -2, 0)
D. (0,-2,-4)

Answer

To find the points where the line intersects the surface defined by the equation \(x^2 + y^2 + z^2 - 20 = 0\), we will follow these steps:

### Step 1: Parameterize the Line
The line is given by the equation:
\[ \frac{x-1}{1} = \frac{y-1}{-1} = \frac{z+3}{1} \]
Let us set this equal to a parameter \(k\):
\[ x - 1 = k \implies x = k + 1 \]
\[ y - 1 = -k \implies y = 1 - k \]
\[ z + 3 = k \implies z = k - 3 \]

### Step 2: Substitute into the Surface Equation
Now we substitute \(x\), \(y\), and \(z\) into the surface equation \(x^2 + y^2 + z^2 - 20 = 0\):
\[ (k + 1)^2 + (1 - k)^2 + (k - 3)^2 - 20 = 0 \]

### Step 3: Expand the Equation
Expanding each term:
\[ (k + 1)^2 = k^2 + 2k + 1 \]
\[ (1 - k)^2 = 1 - 2k + k^2 \]
\[ (k - 3)^2 = k^2 - 6k + 9 \]
Now combine these:
\[ (k^2 + 2k + 1) + (1 - 2k + k^2) + (k^2 - 6k + 9) - 20 = 0 \]
This simplifies to:
\[ 3k^2 - 6k + 11 - 20 = 0 \]
\[ 3k^2 - 6k - 9 = 0 \]

### Step 4: Simplify the Quadratic Equation
Dividing the entire equation by 3:
\[ k^2 - 2k - 3 = 0 \]

### Step 5: Factor the Quadratic
Factoring the quadratic:
\[ (k - 3)(k + 1) = 0 \]
Thus, the solutions for \(k\) are:
\[ k = 3 \quad \text{or} \quad k = -1 \]

### Step 6: Find the Points of Intersection
Now we substitute back to find the points corresponding to \(k = 3\) and \(k = -1\).

For \(k = 3\):
\[ x = 3 + 1 = 4, \quad y = 1 - 3 = -2, \quad z = 3 - 3 = 0 \]
So one point is \((4, -2, 0)\).

For \(k = -1\):
\[ x = -1 + 1 = 0, \quad y = 1 - (-1) = 2, \quad z = -1 - 3 = -4 \]
So another point is \((0, 2, -4)\).

### Final Answer
The points where the line intersects the surface are:
\[ (4, -2, 0) \quad \text{and} \quad (0, 2, -4) \]