If a, b, c, are three rational numbers, $\mathrm{a}-7\mathrm{b}+8\mathrm{c}=4$ and $8\mathrm{a}+4\mathrm{b}-\mathrm{c}=7$,  then the value of ${\mathrm{a}}^2-{\mathrm{b}}^2+{\mathrm{c}}^2=$

1) $a-7b+8c=4$
$\mathrm{a}+8\mathrm{c}=4+7\mathrm{b}$ (sq on both sides )
$\Rightarrow a^2+16ac+64c^2=16+56b+49b^2\to \left(1\right)$
ii) $$\begin{gathered} 8a+4b-c=7 \\ (8a-c)=(7-4b)(sq on both sides) \end{gathered}$$
$64{\mathrm{a}}^2+{\mathrm{c}}^2+16\mathrm{ac}=49+16{\mathrm{b}}^2-56\mathrm{b}\to \left(2\right)$
Adding (1) and (2) 
$$\begin{gathered} {\mathrm{a}}^2+16\mathrm{ac}+64{\mathrm{c}}^2=16-56\mathrm{b}+49{\mathrm{b}}^2 \\ \underset{¯}{64{\mathrm{a}}^2-16\mathrm{ac}+{\mathrm{c}}^2=49-56\mathrm{b}+16{\mathrm{b}}^2} \\ \Rightarrow 65{\mathrm{a}}^2+65{\mathrm{c}}^2=65+65{\mathrm{b}}^2 \\ \Rightarrow 65\left({\mathrm{a}}^2+{\mathrm{c}}^2\right)=65\left(1+{\mathrm{b}}^2\right) \\ \Rightarrow {\mathrm{a}}^2+{\mathrm{c}}^2=1+{\mathrm{b}}^2 \\ \therefore {\mathrm{a}}^2-{\mathrm{b}}^2+{\mathrm{c}}^2=1 \end{gathered}$$