Let the point A be the foot of perpendicular drawn from the point P(a, b, 0) on the line x-1 2 = y-2 1 = z-α 3. If the midpoint of the line segment…

Let $P = (a, b, 0)$P=(a,b,0) and the midpoint of $PA$PA be $M = \left(0, \dfrac{3}{4}, -\dfrac{1}{4}\right)$M=(0,43​,−41​).

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

$$A = \left(-a, \dfrac{3}{2} - b, -\dfrac{1}{2}\right)$$A=(−a,23​−b,−21​)

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

$$\dfrac{-a-1}{2} = \lambda \Rightarrow a = -2\lambda - 1$$2−a−1​=λ⇒a=−2λ−1

$$\dfrac{\dfrac{3}{2}-b-2}{1} = \lambda \Rightarrow b = -\lambda - \dfrac{1}{2}$$123​−b−2​=λ⇒b=−λ−21​

$$\dfrac{-\dfrac{1}{2}-\alpha}{3} = \lambda \Rightarrow \alpha = -3\lambda - \dfrac{1}{2}$$3−21​−α​=λ⇒α=−3λ−21​

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

$$\vec{PA} = \left(-2a, \dfrac{3}{2} - 2b, -\dfrac{1}{2}\right)$$PA=(−2a,23​−2b,−21​)

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

$$2(-2a) + 1\left(\dfrac{3}{2} - 2b\right) + 3\left(-\dfrac{1}{2}\right) = 0$$2(−2a)+1(23​−2b)+3(−21​)=0

$$-4a + \dfrac{3}{2} - 2b - \dfrac{3}{2} = 0$$−4a+23​−2b−23​=0

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

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

$$2(-2\lambda - 1) + \left(-\lambda - \dfrac{1}{2}\right) = 0$$2(−2λ−1)+(−λ−21​)=0

$$-4\lambda - 2 - \lambda - \dfrac{1}{2} = 0$$−4λ−2−λ−21​=0

$$-5\lambda - \dfrac{5}{2} = 0 \Rightarrow \lambda = -\dfrac{1}{2}$$−5λ−25​=0⇒λ=−21​

Now, substituting $\lambda = -\dfrac{1}{2}$λ=−21​ back to find $a, b,$a,b, and $\alpha$α:

$$a = -2\left(-\dfrac{1}{2}\right) - 1 = 0$$a=−2(−21​)−1=0

$$b = -\left(-\dfrac{1}{2}\right) - \dfrac{1}{2} = 0$$b=−(−21​)−21​=0

$$\alpha = -3\left(-\dfrac{1}{2}\right) - \dfrac{1}{2} = \dfrac{3}{2} - \dfrac{1}{2} = 1$$α=−3(−21​)−21​=23​−21​=1

Finally, calculating the required value:

$$a^2 + b^2 + \alpha^2 = 0^2 + 0^2 + 1^2 = 1$$a2+b2+α2=02+02+12=1

Answer: $1$1