Let the point A be the foot of perpendicular drawn from the point $P(a, b, 0)$ on the line $\dfrac{x-1}{2} = \dfrac{y-2}{1} = \dfrac{z-\alpha}{3}$. If the midpoint of the line segment PA is $\left(0, \dfrac{3}{4}, \dfrac{-1}{4}\right)$, then the value of $a^2 + b^2 + \alpha^2$ is equal to:

Let $P = (a, b, 0)$ and the midpoint of $PA$ be $M = \left(0, \dfrac{3}{4}, -\dfrac{1}{4}\right)$.

Using the midpoint formula, the coordinates of $A$ are given by $2M - P$:

$A = \left(-a, \dfrac{3}{2} - b, -\dfrac{1}{2}\right)$

Since $A$ lies on the given line $\dfrac{x-1}{2} = \dfrac{y-2}{1} = \dfrac{z-\alpha}{3} = \lambda$, we can write:

$\dfrac{-a-1}{2} = \lambda \Rightarrow a = -2\lambda - 1$

$\dfrac{\dfrac{3}{2}-b-2}{1} = \lambda \Rightarrow b = -\lambda - \dfrac{1}{2}$

$\dfrac{-\dfrac{1}{2}-\alpha}{3} = \lambda \Rightarrow \alpha = -3\lambda - \dfrac{1}{2}$

The direction ratios of the line segment $PA$ are the differences in coordinates of $A$ and $P$:

$\vec{PA} = \left(-2a, \dfrac{3}{2} - 2b, -\dfrac{1}{2}\right)$

Since $PA$ is perpendicular to the given line, the dot product of their direction ratios must be zero. The direction ratios of the line are $(2, 1, 3)$.

$2(-2a) + 1\left(\dfrac{3}{2} - 2b\right) + 3\left(-\dfrac{1}{2}\right) = 0$

$-4a + \dfrac{3}{2} - 2b - \dfrac{3}{2} = 0$

$-4a - 2b = 0 \Rightarrow 2a + b = 0$

Substituting the expressions for $a$ and $b$ in terms of $\lambda$:

$2(-2\lambda - 1) + \left(-\lambda - \dfrac{1}{2}\right) = 0$

$-4\lambda - 2 - \lambda - \dfrac{1}{2} = 0$

$-5\lambda - \dfrac{5}{2} = 0 \Rightarrow \lambda = -\dfrac{1}{2}$

Now, substituting $\lambda = -\dfrac{1}{2}$ back to find $a, b,$ and $\alpha$:

$a = -2\left(-\dfrac{1}{2}\right) - 1 = 0$

$b = -\left(-\dfrac{1}{2}\right) - \dfrac{1}{2} = 0$

$\alpha = -3\left(-\dfrac{1}{2}\right) - \dfrac{1}{2} = \dfrac{3}{2} - \dfrac{1}{2} = 1$

Finally, calculating the required value:

$a^2 + b^2 + \alpha^2 = 0^2 + 0^2 + 1^2 = 1$

Answer: $1$