Answer
$Σresidual^2=1$
Work Step by Step
Since $ŷ =2x−2$
$Σresidual^2=Σ(y_i-ŷ _i)^2=Σ[y_i-(2x_i-2)]^2=Σ(y_i-2x_i+2)^2$
$Σresidual^2=(4-2\times3+2)^2+(6-2\times4+2)^2+(7-2\times5+2)^2+(12-2\times7+2)^2+(14-2\times8+2)^2$
$Σresidual^2=1$