If the foot of the perpendicular drawn from the point $(0,2,1)$ on a line passing through $(\alpha, 5,1)$ is $\left(\frac{5}{3}, \frac{7}{3}, \frac{15}{i}\right)$, then $\alpha$ is equal to _____

Solution

It is given that point $Q\left(\frac{5}{3}, \frac{7}{3}, \frac{15}{4}\right)$ is foot of perpendicular of point $P(0,2,1)$ to a line passes through point $A(\alpha, 5,1)$. So, $P Q \perp A Q$.
$\because$ Direction ratios of line segments $P Q$ is $\left(\frac{5}{3}, \frac{1}{3}, \frac{11}{4}\right)$ and direction ratios of line $A Q$ is $\left(\alpha-\frac{5}{3}, \frac{8}{3},-\frac{11}{4}\right)$. $\begin{array}{cc} \therefore & \frac{5}{3}\left(\alpha-\frac{5}{3}\right)+\frac{1}{3} \times \frac{8}{3}+\frac{11}{4} \times\left(-\frac{11}{4}\right)=0 \\ \Rightarrow & 15 \alpha-25+8+\frac{99(-11)}{16}=0 \\ \Rightarrow & 15 \alpha-17=\frac{99 \times 11}{16} \\ \Rightarrow & 5 \alpha=\frac{1089-817}{16} \\ \Rightarrow & 5 \alpha=54.466 \Rightarrow \alpha=10.893 \end{array}$

Hence, the correct answer is 10.893.