The length of the perpendicular from the point $P(1,-1,2)$ to the given line $\dfrac{x+1}{2}=\dfrac{y-2}{-3}=\dfrac{z+2}{4}$ is

Let the line be $L: \dfrac{x+1}{2} = \dfrac{y-2}{-3} = \dfrac{z+2}{4} = \lambda$.

Any point $Q$ on the line $L$ is given by $Q(2\lambda - 1, -3\lambda + 2, 4\lambda - 2)$.

The vector $\overrightarrow{PQ}$ is given by $(2\lambda - 1 - 1, -3\lambda + 2 - (-1), 4\lambda - 2 - 2) = (2\lambda - 2, -3\lambda + 3, 4\lambda - 4)$.

Since $\overrightarrow{PQ}$ is perpendicular to the direction vector of the line $\vec{v} = (2, -3, 4)$, their dot product must be zero:

$(2\lambda - 2)(2) + (-3\lambda + 3)(-3) + (4\lambda - 4)(4) = 0$

$4\lambda - 4 + 9\lambda - 9 + 16\lambda - 16 = 0$

$29\lambda - 29 = 0 \Rightarrow \lambda = 1$.

Substituting $\lambda = 1$ into the coordinates of $Q$, we get $Q(2(1) - 1, -3(1) + 2, 4(1) - 2) = Q(1, -1, 2)$.

Since the point $P(1, -1, 2)$ lies on the line, the length of the perpendicular from $P$ to the line is 0.

Answer: 0 units