If the line $\frac{x+1}{1}=\frac{y-k}{11}=\frac{z-4}{-5}$ lies in the plane $2 x+p y+7 z-41=0$ which is perpendicular to the plane $x+4 \mathrm{y}-2 \mathrm{z}+13=0$ then $\mathrm{k}=$

Given: Line $L: \frac{x+1}{1} = \frac{y-k}{11} = \frac{z-4}{-5}$ with direction vector $\vec{d} = (1, 11, -5)$ and point $P_0(-1, k, 4)$.
Plane $P_1: 2x + py + 7z - 41 = 0$ with normal vector $\vec{n_1} = (2, p, 7)$.
Plane $P_2: x + 4y - 2z + 13 = 0$ with normal vector $\vec{n_2} = (1, 4, -2)$.

Since $P_1 \perp P_2$, their normal vectors are perpendicular: $\vec{n_1} \cdot \vec{n_2} = 0$.

$(2, p, 7) \cdot (1, 4, -2) = 2 + 4p - 14 = 0 \Rightarrow 4p = 12 \Rightarrow p = 3$.

Substituting $p = 3$ into $P_1$: $2x + 3y + 7z - 41 = 0$.

For $L$ to lie in $P_1$, point $P_0$ must satisfy the plane equation:
$2(-1) + 3k + 7(4) - 41 = -2 + 3k + 28 - 41 = 3k - 15 = 0 \Rightarrow k = 5$.

(Verification: The direction vector $\vec{d}$ is perpendicular to $\vec{n_1}$: $\vec{d} \cdot \vec{n_1} = 1\cdot2 + 11\cdot3 + (-5)\cdot7 = 2 + 33 - 35 = 0$, confirming the result.)

Thus, $k = \boxed{5}$.