The point of intersection of the lines $\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}$ and $\dfrac{x-4}{5}=\dfrac{y-1}{2}=z$ is

Let the first line be $\dfrac{x-1}{2} = \dfrac{y-2}{3} = \dfrac{z-3}{4} = s$. Then the coordinates of any point on this line are $(2s+1, 3s+2, 4s+3)$.

Let the second line be $\dfrac{x-4}{5} = \dfrac{y-1}{2} = \dfrac{z}{1} = t$. Then the coordinates of any point on this line are $(5t+4, 2t+1, t)$.

For the lines to intersect, there must exist values of $s$ and $t$ such that the coordinates are equal:
$2s+1 = 5t+4 \Rightarrow 2s - 5t = 3$
$3s+2 = 2t+1 \Rightarrow 3s - 2t = -1$
$4s+3 = t$

Substituting $t = 4s+3$ into the first equation:
$2s - 5(4s+3) = 3 \Rightarrow 2s - 20s - 15 = 3 \Rightarrow -18s = 18 \Rightarrow s = -1$.

Substituting $s = -1$ into $t = 4s+3$:
$t = 4(-1) + 3 = -1$.

Checking these values in the second equation:
$3(-1) - 2(-1) = -3 + 2 = -1$. This matches the equation $3s - 2t = -1$.

The point of intersection is obtained by substituting $s = -1$ into the first line coordinates:
$x = 2(-1) + 1 = -1$
$y = 3(-1) + 2 = -1$
$z = 4(-1) + 3 = -1$.

The point of intersection is $(-1, -1, -1)$.

Answer: $(-1,-1,-1)$