The line `(x-2)/3=(y+1)/2=(z-1)/1` intersects the curve `x y=c^(2),z=0` then `c` is equal to a. `+-1` b. `+-1//3` c. `+-sqrt(5)` d. none of these

Options

A. `ne1`
B. `+-1//3`
C. `+-sqrt5`
D. none of these

Explanation

To solve the problem, we need to find the value of \( c \) such that the line given by

\[ \frac{x-2}{3} = \frac{y+1}{2} = \frac{z-1}{1} \]

intersects the curve defined by

\[ xy = c^2 \quad \text{and} \quad z = 0. \]

### Step 1: Substitute \( z = 0 \) in the line equation

Since the curve has \( z = 0 \), we can substitute \( z = 0 \) into the line equation:

\[ \frac{x-2}{3} = \frac{y+1}{2} = \frac{0-1}{1} = -1. \]

### Step 2: Solve for \( x \)

Now we can equate \( \frac{x-2}{3} \) to \(-1\):

\[ \frac{x-2}{3} = -1. \]

Multiplying both sides by 3 gives:

\[ x - 2 = -3. \]

Adding 2 to both sides results in:

\[ x = -1. \]

### Step 3: Solve for \( y \)

Next, we equate \( \frac{y+1}{2} \) to \(-1\):

\[ \frac{y+1}{2} = -1. \]

Multiplying both sides by 2 gives:

\[ y + 1 = -2. \]

Subtracting 1 from both sides results in:

\[ y = -3. \]

### Step 4: Substitute \( x \) and \( y \) into the curve equation

Now we have the coordinates \( (x, y) = (-1, -3) \). We substitute these values into the curve equation \( xy = c^2 \):

\[ (-1)(-3) = c^2. \]

This simplifies to:

\[ 3 = c^2. \]

### Step 5: Solve for \( c \)

Taking the square root of both sides gives:

\[ c = \pm \sqrt{3}. \]

### Conclusion

Since \( \pm \sqrt{3} \) is not listed in the options provided, the correct answer is:

**d. none of these.**