Let  the equation of the plane P containing the line $\mathrm{x}+10=\frac{8-\mathrm{y}}{2}=\mathrm{z}\text{ be }\mathrm{ax}+\mathrm{by}+3\mathrm{z}=2(\mathrm{a}+\mathrm{b})$ and the distance of the plane P from the point (1, 27, 7) be $c$. Then ${\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2$ is equal to______.

A point on the line is (-10, 8, 0)
(-10, 8, 0) lies on P
$\begin{array}{l}\Rightarrow \mathrm{a}(-10)+\mathrm{b}\left(8\right)+0=2(\mathrm{a}+\mathrm{b})\\ \Rightarrow 12\mathrm{a}-6\mathrm{b}=0\Rightarrow 2\mathrm{a}=\mathrm{b}\end{array}$
Drs of the line are 1, -2, 1 
Drs of the normal to P are a, b, 3
$\begin{array}{l}1\cdot \mathrm{a}-2\cdot \mathrm{b}+1.3=0\\ \Rightarrow \mathrm{a}-4\mathrm{a}+3=0\text{ from }\left(1\right)\\ \Rightarrow \mathrm{a}=1,\mathrm{b}=2\\ \text{ P is }\mathrm{x}+2\mathrm{y}+3\mathrm{z}-6=0\\ \mathrm{c}=\frac{|1+2.27+3.7-6|}{\sqrt{1+4+9}}\\ =\frac{70}{\sqrt{14}}=5\sqrt{14}\\ {\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2=355\end{array}$