If the lines $\dfrac{x-1}{2} = \dfrac{y+1}{3} = \dfrac{z-1}{4}$ and $\dfrac{x-2}{1} = \dfrac{y+m}{2} = \dfrac{z-2}{1}$ intersect each other, then the value of $m$ is....

Let the given lines be $L_1$ and $L_2$.

Any point on $L_1$ is given by $(2r_1 + 1, 3r_1 - 1, 4r_1 + 1)$.

Any point on $L_2$ is given by $(r_2 + 2, 2r_2 - m, r_2 + 2)$.

For the lines to intersect, these points must coincide for some values of $r_1$ and $r_2$. Equating the $x$, $y$, and $z$ coordinates:

$2r_1 + 1 = r_2 + 2 \Rightarrow 2r_1 - r_2 = 1$

$3r_1 - 1 = 2r_2 - m \Rightarrow 3r_1 - 2r_2 = 1 - m$

$4r_1 + 1 = r_2 + 2 \Rightarrow 4r_1 - r_2 = 1$

Solving the first and third equations:

$4r_1 - r_2 = 1$

$2r_1 - r_2 = 1$

Subtracting the two equations gives $2r_1 = 0 \Rightarrow r_1 = 0$.

Substituting $r_1 = 0$ in $2r_1 - r_2 = 1$ gives $r_2 = -1$.

Substituting $r_1 = 0$ and $r_2 = -1$ in the second equation:

$3(0) - 2(-1) = 1 - m$

$2 = 1 - m$

$m = -1$

Answer: $-1$