If the foot of the perpendicular drawn from the point $(1,0,3)$ on $a$ line passing through $(\alpha, 7,1)$ is $\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right)$, then $\alpha$ is equal to__________.
Since, $P Q$ is perpendicular to $\mathrm{L}$
$Q\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right)$
$\therefore \quad\left(1-\frac{5}{3}\right)\left(\alpha-\frac{5}{3}\right)+\left(\frac{-7}{3}\right)\left(7-\frac{7}{3}\right)$
$+\left(3-\frac{17}{3}\right)\left(1-\frac{17}{3}\right)=0$
$\Rightarrow \frac{-2 \alpha}{3}+\frac{10}{9}-\frac{98}{9}+\frac{112}{9}=0$
$\Rightarrow \quad \frac{-2 \alpha}{3}+\frac{10}{9}-\frac{98}{9}+\frac{112}{9}=0$
$\Rightarrow \quad \frac{2}{3}=\frac{24}{9} \Rightarrow \alpha=4$
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