If the length of the perpendicular drawn from the point P(a, 4, 2), a > 0 on the line $\frac{\mathrm{x}+1}{2}=\frac{\mathrm{y}-3}{3}=\frac{\mathrm{z}-1}{-1}$ is $2\sqrt{6}$ units and $\mathrm{Q}\left({\mathrm{\alpha }}_1, {\mathrm{\alpha }}_2, {\mathrm{\alpha }}_3\right)$ is the image of the point P in this line, then $\mathrm{a}+\sum _{\mathrm{i}=1}^3 {\mathrm{\alpha }}_{\mathrm{i}}$ is equal to :

$$\begin{gathered} \frac{\mathrm{x}+1}{2}=\frac{\mathrm{y}-3}{3}=\frac{\mathrm{z}-1}{-1}=\lambda \\ \Rightarrow x=2\lambda -1,y=3\lambda +3,z=-\lambda +1 \end{gathered}$$

$\begin{array}{l}(2\mathrm{\lambda }-1-\mathrm{a})2+(3\mathrm{\lambda }-1)3+(-\mathrm{\lambda }-1)(-1)=0\\ \Rightarrow 4\mathrm{\lambda }-2-2\mathrm{a}+9\mathrm{\lambda }-3+\mathrm{\lambda }+1=0\\ \Rightarrow 14\mathrm{\lambda }-4-2\mathrm{a}=0\\ \Rightarrow 7\mathrm{\lambda }-2-\mathrm{a}=0\end{array}$

and,

$\begin{array}{l}(2\mathrm{\lambda }-1-\mathrm{a}{)}^2+(3\mathrm{\lambda }-1{)}^2+(\mathrm{\lambda }+1{)}^2=24\\ \Rightarrow (5\mathrm{\lambda }-1{)}^2+(3\mathrm{\lambda }-1{)}^2+(\mathrm{\lambda }+1{)}^2=24\\ \Rightarrow 35{\mathrm{\lambda }}^2-14\mathrm{\lambda }-21=0\\ \Rightarrow (\mathrm{\lambda }-1)(35\mathrm{\lambda }+21)=0\end{array}$

For, $\mathrm{\lambda }=1 \Rightarrow \mathrm{a}=5$

Let $\left({\mathrm{\alpha }}_1, {\mathrm{\alpha }}_2, {\mathrm{\alpha }}_3\right)$ be reflection of point P

$$\begin{gathered} {\mathrm{\alpha }}_1+5=2 {\mathrm{\alpha }}_2+4=12 {\mathrm{\alpha }}_3+2=0 \\ {\mathrm{\alpha }}_1=-3 {\mathrm{\alpha }}_2=8 {\mathrm{\alpha }}_3=-2 \\ \mathrm{a}+{\mathrm{\alpha }}_1+{\mathrm{\alpha }}_2+{\mathrm{\alpha }}_3=8 \end{gathered}$$