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The problem provided asks to determine the intersection point \( p \) of a line \( l \) and a plane \( V \) given by the equation \( 4x + 2y + 3z = -10 \).

### Step 1: Parametric Equation of the Line
The line \( l \) passes through the points \( (4, -4, -4) \) and \( (-5, 1, 4) \). First, let's find the direction vector of the line by subtracting the coordinates of these points:

\[
\vec{d} = (-5 - 4, 1 + 4, 4 + 4) = (-9, 5, 8)
\]

Thus, the direction vector is \( \vec{d} = (-9, 5, 8) \).

The parametric equation of the line is then:

\[
(x, y, z) = (4, -4, -4) + t(-9, 5, 8)
\]
where \( t \) is a parameter. This gives the system of parametric equations:

\[
x = 4 - 9t, \quad y = -4 + 5t, \quad z = -4 + 8t
\]

### Step 2: Substituting into the Plane Equation
Now, substitute these parametric equations into the plane equation \( 4x + 2y + 3z = -10 \):

\[
4(4 - 9t) + 2(-4 + 5t) + 3(-4 + 8t) = -10
\]

Simplifying:

\[
16 - 36t - 8 + 10t - 12 + 24t = -10
\]

\[
(16 - 8 - 12) + (-36t + 10t + 24t) = -10
\]

\[
-4 + (-2t) = -10
\]

\[
-2t = -6 \quad \Rightarrow \quad t = 3
\]

### Step 3: Find the Coordinates of the Intersection Point
Now that we have \( t = 3 \), substitute this value back into the parametric equations for \( x \), \( y \), and \( z \):

\[
x = 4 - 9(3) = 4 - 27 = -23
\]
\[
y = -4 + 5(3) = -4 + 15 = 11
\]
\[
z = -4 + 8(3) = -4 + 24 = 20
\]

Thus, the coordinates of the intersection point are \( (-23, 11, 20) \).

Would you like more details on any part of the solution or have any questions?

### Related Questions:
1. How can you find the parametric equations of a line given two points?
2. How do you check if a point lies on a given plane?
3. What are the conditions for a line and a plane to be parallel or coincide?
4. How can you derive the equation of a plane given a point and a normal vector?
5. Can you have more than one intersection point between a line and a plane?

### Tip:
When solving for intersections, it's helpful to express the line in parametric form and substitute into the plane's equation directly.

Determine the intersection point p of the line l and the plane V given by the equation 4x + 2y + 3z = -10, where the line passes through the points (4, -4, -4) and (-5, 1, 4).

Learn how to find the intersection point of a line passing through (4, -4, -4) and (-5, 1, 4) with a plane given by 4x + 2y + 3z = -10 using parametric equations. This step-by-step solution covers key concepts in 3D analytic geometry.

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