Mathematics - Vector Question with Solution | TestHub

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