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

Point P=(1,0,3)P = (1, 0, 3)P=(1,0,3)

Line passes through A=(α,7,1)A = ($$\alpha\ $$, 7, 1)A=(α,7,1)

Foot of perpendicular from P to the line is

$$F=\ \left(\frac{5}{3},\frac{7}{3},\frac{17}{3}\right)$$

Now the main ideas :

Dot product of perpendicular vectors is = 0

Line passes through A and F, its direction vector is:
$$\vec{AF}$$= F - A

$$\vec{PF}$$ = F - P 

  $$\vec{PF}$$ = $$\left(\frac{5}{3}-\alpha\ ,\ \frac{7}{3}-7,\ \frac{17}{3}-1\right)$$

                = $$\left(\frac{5}{3}-\alpha\ ,\ -\frac{14}{3},\ \frac{14}{3}\right)$$

Now use perpendicular condition 

     $$\vec{PF}$$ . $$\vec{AF}$$ = 0

     $$\left(\frac{2}{3},\ \frac{7}{3},\ \frac{8}{3}\right)\ .\ \left(\frac{5}{3}-\alpha\ ,\ -\frac{14}{3},\ \frac{14}{3}\right)\ \ =\ 0$$

Compute dot product and equal it to Zero

$$\frac{2}{3}\left(\frac{5}{3}-\alpha\ \right)+\frac{7}{3}\left(-\frac{14}{3}\right)+\frac{8}{3}\left(\frac{14}{3}\right)\ \ =\ 0$$

$$\frac{10}{9}-\frac{2\alpha}{3}-\frac{98}{9}+\frac{112}{9}\ =\ 0$$

$$\frac{8}{3}\ -\ \frac{2\alpha}{3}\ =\ 0$$

$$8\ -\ 2\alpha\ \ =\ 0$$

$$2\alpha\ \ =\ 8$$

$$\alpha\ \ =\ 4$$

                                                Hence the answer is $$\alpha\ \ =4$$