Calculate Sum of Squares of Coordinates from Midpoint of Perpendicular Segment

To find the value of $a^2 + b^2 + \alpha^2$, we express the coordinates of the foot of the perpendicular $A$ in terms of $P(a,b,0)$ using the given midpoint $\left(0, \frac{3}{4}, -\frac{1}{4}\right)$, which gives $A\left(-a, \frac{3}{2}-b, -\frac{1}{2}\right)$.

Since $A$ lies on the line $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - \alpha}{3}$ and the vector $\vec{PA}$ is perpendicular to the line's direction vector $\vec{d} = (2, 1, 3)$, setting up the position equations and orthogonality condition $\vec{PA} \cdot \vec{d} = 0$ yields a system of equations in $a$, $b$, and $\alpha$. Solving this system gives $a = 0$, $b = 0$, and $\alpha = 1$, which leads to $a^2 + b^2 + \alpha^2 = 1$.