Explanation
\begin{aligned}
& \mathrm{Given:}1+x+x^{2}=0 \\
& \therefore\alpha=\omega,\omega^{2} \\
& \mathrm{Now},\left(1+\alpha\right)^{7}=a+b\alpha+c\alpha^{2} \\
& \Rightarrow(1+\omega)^{7}=a+b\omega+c\alpha^{2} \\
& \Rightarrow-\omega^{14}=a+b\omega+c\omega^{2} \\
& \Rightarrow-\omega^{2}=a+b\omega+c\omega^{2} \\
& \Rightarrow a+b\omega+(c+1)\omega^{2}=0 \\
& \Rightarrow a+b\left(\frac{-1}{2}+\frac{\sqrt{3}}{2}i\right)+(c+1)\left(\frac{-1}{2}-\frac{\sqrt{3}}{2}i\right)=0 \\
& \Rightarrow a-\frac{b}{2}-\frac{c+1}{2}=0\Rightarrow2a-b-c-1=0 \\
& & \frac{\sqrt{3}}{2}b-\frac{\sqrt{3}}{2}(c+1)=0 \\
& \Rightarrow b-c-1=0 \\
& \mathrm{So},a=b \\
& \therefore5(3a-2b-c)=5(b-c)=5
\end{aligned}
