To find the intersection, equate the parametric equations of the lines:

\(\begin{pmatrix} 5 + s \\ 1 - s \\ -4 + 3s \end{pmatrix} = \begin{pmatrix} p + 2t \\ 4 + 5t \\ -2 - 4t \end{pmatrix}\).

Equating components, we get:

1. \(5 + s = p + 2t\)

2. \(1 - s = 4 + 5t\)

3. \(-4 + 3s = -2 - 4t\)

From equation 2: \(s = -3 - 5t\).

Substitute \(s = -3 - 5t\) into equation 3:

\(-4 + 3(-3 - 5t) = -2 - 4t\)

\(-4 - 9 - 15t = -2 - 4t\)

\(-13 - 15t = -2 - 4t\)

\(-11 = 11t\)

\(t = -1\).

Substitute \(t = -1\) into \(s = -3 - 5t\):

\(s = -3 - 5(-1) = 2\).

Substitute \(s = 2\) and \(t = -1\) into equation 1:

\(5 + 2 = p + 2(-1)\)

\(7 = p - 2\)

\(p = 9\).

Thus, the value of \(p\) is 9. The intersection point is:

\(\begin{pmatrix} 5 + 2 \\ 1 - 2 \\ -4 + 3(2) \end{pmatrix} = \begin{pmatrix} 7 \\ -1 \\ 2 \end{pmatrix}\).