The d.r.s. of the normal to the plane are $1,-1,2$.
$\therefore \quad$ The equation of line PM is
$\begin{aligned}
& \frac{x-2}{1}=\frac{y-4}{-1}=\frac{\mathrm{z}+1}{2}=\lambda(\text { say }) \\
& \Rightarrow x=\lambda+2, y=-\lambda+4, \mathrm{z}=2 \lambda-1 \\
& \text { Let } \mathrm{M} \equiv(\lambda+2,-\lambda+4,2 \lambda-1)
\end{aligned}$
$\begin{array}{ll}
\therefore \quad & \text { Equation of plane becomes } \\
& 1(\lambda+2)-1(-\lambda+4)+2(2 \lambda-1)-2=0 \\
& \Rightarrow \lambda=1 \\
\therefore \quad & M \equiv(3,3,1)
\end{array}$
Since $M$ is the mid-point of $P Q$.
$\begin{aligned}
& \therefore \quad \frac{2+\mathrm{a}}{2}=3, \frac{4+\mathrm{b}}{2}=3, \frac{-1+\mathrm{c}}{2}=1 \\
& \Rightarrow \mathrm{a}=4, \mathrm{~b}=2, \mathrm{c}=3 \\
& \Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}=9
\end{aligned}$