If the image of the point $\mathrm{P}(1,0,3)$ in the line joining the points $\mathrm{A}(4,7,1)$ and $\mathrm{B}(3,5,3)$ is $\mathrm{Q}(\alpha, \beta, \gamma)$, then $\alpha+\beta+\gamma$ is equal to

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{P}(1,0,3) \ & \mathrm{A}(4,7,1), \mathrm{B}(3,5,3) \ & 	ext { Line } \mathrm{AB} \Rightarrow \frac{\mathrm{x}-3}{1}=\frac{\mathrm{y}-5}{2}=\frac{\mathrm{z}-3}{-2}=\lambda \end{aligned}$ Let foot of perpendicular of P on AB be $\begin{aligned} & \mathrm{R} \equiv(\lambda+3,2 \lambda+5,-2 \lambda+3) \ & \Rightarrow(\lambda+3-1)(1)+(2 \lambda+5-0)(2)+(-2 \lambda+3-3) \ & (-2)=0 \ & \Rightarrow \lambda+2+4 \lambda+10+4 \lambda=0 \ & \Rightarrow \lambda=-\frac{4}{3} \ & \Rightarrow \mathrm{R} \equiv\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}ight) \ & \mathrm{Q} \equiv\left(\frac{10}{3}-1, \frac{14}{3}-0, \frac{34}{3}-3ight) \equiv\left(\frac{7}{3}, \frac{14}{3}, \frac{25}{3}ight) \ & \Rightarrow \alpha+\beta+\gamma=\frac{7+14+25}{3}=\frac{46}{3}\end{aligned}$ *

If the image of the point $\mathrm{P}(1,0,3)$ in the line joining the points $\mathrm{A}(4,7,1)$ and…

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