If the foot of the perpendicular drawn from the point $(2,0,1)$ on a line passing through $(\alpha, 5,1)$ is $\left(\frac{7}{3}, \frac{5}{3}, \frac{15}{4}\right)$, then $\alpha$ is equal to_____

Solution

It is given that $Q\left(\frac{7}{3}, \frac{5}{3}, \frac{15}{4}\right)$ is foot of perpendicular of point $P(2,0,1)$ 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 $\left(\frac{7}{3}, \frac{5}{3}, \frac{15}{4}\right)$ and direction ratios of line $A Q$ is $\left(\alpha-\frac{7}{3}, \frac{10}{3},-\frac{11}{4}\right)$ Since, $P Q$ and $A Q$ is perpendicular to each other, $\text { so } P Q \cdot A Q=0$ $\frac{7}{3}\left(\alpha-\frac{7}{3}\right)+\frac{5}{3} \times \frac{10}{3}+\frac{15}{4} \times\left(-\frac{11}{4}\right)=0$ $\Rightarrow \quad \frac{7(3 \alpha-7)}{9}+\frac{50}{9}-\frac{165}{16}=0$ $\begin{aligned} & \Rightarrow \quad \frac{336 \alpha-784+800-1485}{144}=0 \\ & \Rightarrow \quad 336 \alpha=1469 \Rightarrow \alpha=\frac{1469}{336}=4.372\end{aligned}$

Hence, the correct answer is 4.372.