The equation of the plane containing the line $\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-4}{-2}$ and the point $(0,5,0)$ is

A plane containing a given line and a point not on the line is determined by the direction vector of the line and the vector from a point on the line to the given point. Using the symmetric form $\frac{x-2}{3} = \frac{y+1}{2} = \frac{z-4}{-2}$, the point $P_1 = (2, -1, 4)$ lies on the line, and the direction vector is $\mathrm{d} = (3, 2, -2)$.

The given point is $P_2 = (0, 5, 0)$, so $\mathrm{P_1P_2} = (-2, 6, -4)$. The normal vector $\mathrm{n}$ to the plane is the cross product $\mathrm{d} \times \mathrm{P_1P_2}$.

$n_x = (2)(-4) - (-2)(6) = 4$, $n_y = -[(3)(-4) - (-2)(-2)] = -[-12 - 4] = 16$, $n_z = (3)(6) - (2)(-2) = 18 + 4 = 22$, giving $\mathrm{n} = (4, 16, 22)$. Use a simpler normal vector: $(2, 8, 11)$ after division by 2.

Using $\mathrm{n} = (2, 8, 11)$ and point $P_2 = (0, 5, 0)$, the plane equation is $2(x - 0) + 8(y - 5) + 11(z - 0) = 0$, simplifying to $2x + 8y + 11z - 40 = 0$.

Final Answer: \boxed{\text{B}}