Let the point A be the foot of the perpendicu | vedclass

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(A) Let the line be $L: \frac{x-1}{2} = \frac{y-2}{1} = \frac{z-\alpha}{3} = k$. Any point $A$ on the line is $(2k+1, k+2, 3k+\alpha)$.The midpoint $M

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(A) Let the line be $L: \frac{x-1}{2} = \frac{y-2}{1} = \frac{z-\alpha}{3} = k$. Any point $A$ on the line is $(2k+1, k+2, 3k+\alpha)$.The midpoint $M$ of $PA$ is given by $(\frac{a+2k+1}{2}, \frac{b+k+2}{2}, \frac{0+3k+\alpha}{2}) = (0, \frac{3}{4}, -\frac{1}{4})$.Equating the coordinates:$a+2k+1 = 0 \implies a = -2k-1$$b+k+2 = 1.5 \implies b = -k-0.5$$3k+\alpha = -0.5 \implies \alpha = -3k-0.5$Since $PA$ is perpendicular to the line with direction vector $\vec{v} = (2, 1, 3)$, the vector $\vec{PA} = (2k+1-a, k+2-b, 3k+\alpha-0)$ must satisfy $\vec{PA} \cdot \vec{v} = 0$.Substituting $a, b, \alpha$ in terms of $k$: $\vec{PA} = (2k+1-(-2k-1), k+2-(-k-0.5), 3k+(-3k-0.5)) = (4k+2, 2k+2.5, -0.5)$.$(4k+2)(2) + (2k+2.5)(1) + (-0.5)(3) = 0 \implies 8k+4+2k+2.5-1.5 = 0 \implies 10k+5 = 0 \implies k = -0.5$.Now, $a = -2(-0.5)-1 = 0$, $b = -(-0.5)-0.5 = 0$, and $\alpha = -3(-0.5)-0.5 = 1$.Thus, $a^2+b^2+\alpha^2 = 0^2+0^2+1^2 = 1$.

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

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