The task involves determining the relationship between a given line and a plane, and finding the intersection point if they intersect.

The equations provided are:

1. The parametric equation of the line:

$\frac{x-7}{5} = \frac{y-4}{1} = \frac{z-5}{4}$

2. The equation of the plane:

$3x - y + 2z - 5 = 0$

Let's first rewrite the parametric form of the line:

$\frac{x-7}{5} = \frac{y-4}{1} = \frac{z-5}{4} = t$

Thus, the parametric equations of the line become:

$x = 5t + 7, \quad y = t + 4, \quad z = 4t + 5$

Now, substitute these expressions into the plane equation $3x - y + 2z - 5 = 0$:

$3(5t + 7) - (t + 4) + 2(4t + 5) - 5 = 0$

Simplifying the equation:

$15t + 21 - t - 4 + 8t + 10 - 5 = 0$

$(15t - t + 8t) + (21 - 4 + 10 - 5) = 0$

$22t + 22 = 0$

Solving for $t$:

$t = -1$

Now, substitute $t = -1$ back into the parametric equations of the line to find the coordinates of the intersection point:

$x = 5(-1) + 7 = 2, \quad y = (-1) + 4 = 3, \quad z = 4(-1) + 5 = 1$

Thus, the coordinates of the intersection point are $(2, 3, 1)$.

Would you like more details or have any questions about the steps?

Here are some related questions you might find interesting:

1. How do you derive parametric equations from the symmetric form of a line?
2. What happens if a line is parallel to a plane?
3. How do you find the angle between a line and a plane?
4. How do you check if a point lies on a plane?
5. How can you calculate the distance between a point and a plane?

Tip: When solving for the intersection of a line and a plane, the parametric form of the line often simplifies the calculation, turning the problem into one of solving for a single variable.