If the length of the perpendicular drawn from the point P(a,4,2), a >0 on the line \[\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}\] is \[2 \sqrt{6}\] units and \[Q\left(\alpha_{1}, \alpha_{2}, \alpha_{3}\right)\] is the image of the point P in this line, then \[\mathrm{a}+\sum\limits_{i=1}^{3} \alpha_{i}\] is equal to :
Solutions · 1
Official solution
\[\because\] PR is perpendicular to given line, so
\[2(2\lambda - 1 - a) + 3(3\lambda - 1) - 1( - \lambda - 1) = 0\]
\[ \Rightarrow a = 7\lambda - 2\]
Now,
\[\because\] \[PR = 2\sqrt 6 \]
\[ \Rightarrow {( - 5\lambda + 1)^2} + {(3\lambda - 1)^2} + {(\lambda + 1)^2} = 24\]
\[ \Rightarrow 5{\lambda ^2} - 2\lambda - 3 = 0 \Rightarrow \lambda = 1\] or \[ - {3 \over 5}\]
\[\because\] a>0 so \[\lambda = 1\] and a=5
Now \[\sum\limits_{i = 1}^3 {{\alpha _i} = 2} \] (Sum of co-ordinate of R) - (Sum of coordinates of P)
=2(7)−11=3
\[a + \sum\limits_{i = 1}^3 {{\alpha _i} = 5 + 3 = 8} \]