The lines `(x-5)/4=(y-7)/4=(z+3)/-5 and (x-8)/7=(y-4)/1=(z-5)/3` are coplanar, intersecting at `(1,3,2)` and the equation of the plane in which they lie is `17x-47y-24z+172=0`.

Answer

To solve the problem, we need to verify that the point of intersection of the two lines lies on the given plane. The lines are given in symmetric form, and we have the equation of the plane.

### Step-by-step Solution:

1. **Identify the Point of Intersection**:
The point of intersection is given as \( P(1, 3, 2) \).

2. **Write Down the Equation of the Plane**:
The equation of the plane is given as:
\[ 17x - 47y - 24z + 172 = 0 \]

3. **Substitute the Point into the Plane Equation**:
We will substitute \( x = 1 \), \( y = 3 \), and \( z = 2 \) into the plane equation:
\[ 17(1) - 47(3) - 24(2) + 172 = 0 \]

4. **Calculate Each Term**:
- Calculate \( 17(1) = 17 \)
- Calculate \( -47(3) = -141 \)
- Calculate \( -24(2) = -48 \)
- The constant term is \( +172 \)

5. **Combine the Results**:
Now, we combine all the calculated terms:
\[ 17 - 141 - 48 + 172 \]

6. **Perform the Arithmetic**:
- First, combine \( 17 - 141 = -124 \)
- Then, combine \( -124 - 48 = -172 \)
- Finally, combine \( -172 + 172 = 0 \)

7. **Conclusion**:
Since the left-hand side equals 0, this confirms that the point \( P(1, 3, 2) \) lies on the plane defined by the equation \( 17x - 47y - 24z + 172 = 0 \). Thus, the lines are coplanar.