If the line `(x-1)/(5)=(y-3)/(2)=(z-3)/(2)` intersects the curve `x^(2)-y^(2)=k^(2), z=0`, then the value of 2k can be equal to

Options

A. `-13`
B. `-(13)/(2)`
C. `(13)/(2)`
D. `(13)/(4)`

Answer

To solve the problem, we need to find the intersection of the given line and the curve, and then determine the value of \(2k\).

### Step-by-Step Solution:

1. **Understand the Line Equation:**
The line is given in the symmetric form:
\[ \frac{x-1}{5} = \frac{y-3}{2} = \frac{z-3}{2} \]
We can express the parameters in terms of a parameter \(t\):
\[ x = 5t + 1, \quad y = 2t + 3, \quad z = 2t + 3 \]

2. **Set \(z = 0\):**
Since we are interested in the intersection with the plane \(z = 0\), we set \(z = 0\):
\[ 2t + 3 = 0 \implies t = -\frac{3}{2} \]

3. **Find \(x\) and \(y\) at \(t = -\frac{3}{2}\):**
Substitute \(t = -\frac{3}{2}\) back into the equations for \(x\) and \(y\):
\[ x = 5\left(-\frac{3}{2}\right) + 1 = -\frac{15}{2} + 1 = -\frac{15}{2} + \frac{2}{2} = -\frac{13}{2} \]
\[ y = 2\left(-\frac{3}{2}\right) + 3 = -3 + 3 = 0 \]

4. **Substitute \(x\) and \(y\) into the Curve Equation:**
The curve is given by:
\[ x^2 - y^2 = k^2 \]
Substitute \(x = -\frac{13}{2}\) and \(y = 0\):
\[ \left(-\frac{13}{2}\right)^2 - 0^2 = k^2 \]
\[ \frac{169}{4} = k^2 \]

5. **Solve for \(k\):**
Taking the square root of both sides:
\[ k = \frac{13}{2} \quad \text{(since \(k\) is positive)} \]

6. **Calculate \(2k\):**
Now, we find \(2k\):
\[ 2k = 2 \times \frac{13}{2} = 13 \]

### Final Answer:
The value of \(2k\) can be equal to \(13\).