A line with direction ratios $1, -1, 2$ intersects the lines $\dfrac{x}{2} = \dfrac{y}{3} = \dfrac{z+1}{3}$ and $\dfrac{x+1}{-1} = \dfrac{y-2}{1} = \dfrac{z}{4}$ at the points $P$ and $Q$, respectively. If the length of the line segment $PQ$ is $\alpha$, then $225\alpha^2$ is equal to:

Let the coordinates of $P$ on the first line be $(2\lambda, 3\lambda, 3\lambda - 1)$.

Let the coordinates of $Q$ on the second line be $(-\mu - 1, \mu + 2, 4\mu)$.

The direction ratios of the line segment $PQ$ are proportional to $1, -1, 2$. Thus, we can write:

$\dfrac{-\mu - 1 - 2\lambda}{1} = \dfrac{\mu + 2 - 3\lambda}{-1} = \dfrac{4\mu - 3\lambda + 1}{2} = k$

This gives the system of equations:

$-\mu - 2\lambda - 1 = k$

$\mu - 3\lambda + 2 = -k$

$4\mu - 3\lambda + 1 = 2k$

Adding the first two equations gives:

$-5\lambda + 1 = 0 \Rightarrow \lambda = \dfrac{1}{5}$

Substituting $\lambda = \dfrac{1}{5}$ into the first and third equations:

$-\mu - \dfrac{7}{5} = k \Rightarrow \mu + k = -\dfrac{7}{5}$

$4\mu + \dfrac{2}{5} = 2k \Rightarrow 2\mu - k = -\dfrac{1}{5}$

Adding these two equations gives:

$3\mu = -\dfrac{8}{5} \Rightarrow \mu = -\dfrac{8}{15}$

Substituting $\mu = -\dfrac{8}{15}$ into $\mu + k = -\dfrac{7}{5}$:

$-\dfrac{8}{15} + k = -\dfrac{21}{15} \Rightarrow k = -\dfrac{13}{15}$

The vector $\overrightarrow{PQ}$ is $(k, -k, 2k)$. The length of the line segment $PQ$ is $\alpha = \sqrt{k^2 + (-k)^2 + (2k)^2} = \sqrt{6k^2}$.

Thus, $\alpha^2 = 6k^2 = 6 \left(-\dfrac{13}{15}\right)^2 = 6 \times \dfrac{169}{225} = \dfrac{1014}{225}$.

Therefore, $225\alpha^2 = 1014$.

Answer: $1014$