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

The given line can be written in parametric form using a parameter $$t$$:

$$x = 1 + 2t,\; y = 2 + t,\; z = \alpha + 3t \quad -(1)$$

Thus the coordinates of the foot of the perpendicular $$A$$ are

$$A\,(1 + 2t,\; 2 + t,\; \alpha + 3t).$$

The point from which the perpendicular is drawn is $$P\,(a,\, b,\, 0).$$

Since the midpoint of $$PA$$ is given as $$M\left(0,\,\frac34,\,-\frac14\right),$$ we use the midpoint formula:

$$\frac{a + (1 + 2t)}{2} = 0,\; \frac{b + (2 + t)}{2} = \frac34,\; \frac{0 + (\alpha + 3t)}{2} = -\frac14.$$

Simplifying each relation:

$$a + 1 + 2t = 0 \;\Rightarrow\; a = -1 - 2t \quad -(2)$$

$$b + 2 + t = \frac32 \;\Rightarrow\; b = -\frac12 - t \quad -(3)$$

$$\alpha + 3t = -\frac12 \;\Rightarrow\; \alpha = -\frac12 - 3t \quad -(4)$$

Let $$\vec{d} = (2,\,1,\,3)$$ be the direction vector of the line. For $$A$$ to be the foot of the perpendicular, $$\overrightarrow{PA} \cdot \vec{d} = 0$$:

$$\bigl((1 + 2t) - a,\; (2 + t) - b,\; (\alpha + 3t) - 0\bigr)\cdot(2,1,3)=0.$$

Substituting $$a,b,\alpha$$ from $$(2)\!-\!(4):$$

$$(1 + 2t - (-1 - 2t),\; 2 + t - (-\tfrac12 - t),\; -\tfrac12) \cdot (2,1,3)=0.$$

Simplify the components:

$$\bigl(2 + 4t,\; \tfrac52 + 2t,\; -\tfrac12\bigr)\cdot(2,1,3)=0.$$

Now compute the dot product:

$$2(2 + 4t) + 1\!\left(\tfrac52 + 2t\right) + 3\!\left(-\tfrac12\right) = 0$$

$$\Rightarrow\; 4 + 8t + \tfrac52 + 2t - \tfrac32 = 0$$

$$\Rightarrow\; 5 + 10t = 0 \;\Longrightarrow\; t = -\tfrac12.$$(5)

Plugging $$t = -\tfrac12$$ into $$(2)\!-\!(4):$$

$$a = -1 - 2\!\left(-\tfrac12\right) = 0,$$

$$b = -\tfrac12 - \left(-\tfrac12\right) = 0,$$

$$\alpha = -\tfrac12 - 3\!\left(-\tfrac12\right) = 1.$$

Therefore,

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

Hence the required value is $$1$$.

Option A which is: $$1$$