To find the perpendicular distance from point P to the line l, we can use the vector approach.

First, find the vector PA where A is a point on the line. Let A be the point corresponding to \(\lambda = 0\), so \(A = \begin{pmatrix} 1 \\ 3 \\ -4 \end{pmatrix}\).

Then, \(\mathbf{PA} = \begin{pmatrix} 1 - (-1) \\ 3 - 4 \\ -4 - 11 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \\ -15 \end{pmatrix}\).

The direction vector of the line is \(\mathbf{d} = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}\).

Use the cross product to find the area of the parallelogram formed by PA and d:

\(\mathbf{PA} \times \mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & -15 \\ 2 & 1 & 3 \end{vmatrix} = \begin{pmatrix} (-1)(3) - (-15)(1) \\ -((2)(3) - (-15)(2)) \\ (2)(1) - (-1)(2) \end{pmatrix} = \begin{pmatrix} 12 \\ -36 \\ 4 \end{pmatrix}\).

The magnitude of this cross product is \(\sqrt{12^2 + (-36)^2 + 4^2} = \sqrt{144 + 1296 + 16} = \sqrt{1456}\).

The magnitude of the direction vector \(\mathbf{d}\) is \(\sqrt{2^2 + 1^2 + 3^2} = \sqrt{14}\).

The perpendicular distance is given by \(\frac{\sqrt{1456}}{\sqrt{14}} = \sqrt{104} \approx 10.2\).