Since, $\alpha, \beta$ are the roots of the equation $a x^2+b x+c=0$.
$\therefore \quad \alpha+\beta=-\frac{b}{a} \quad \alpha \beta=\frac{c}{a}$
The quadratic equation whose roots are $\frac{1-\alpha}{\alpha}$ and $\frac{1-\beta}{\beta}$, is
$\begin{aligned} & x^2-\left(\frac{1-\alpha}{\alpha}+\frac{1-\beta}{\beta}\right) x+\frac{1-\alpha}{\alpha} \cdot \frac{1-\beta}{\beta}=0 \\ & \Rightarrow x^2-\left(\frac{\beta-\alpha \beta+\alpha-\alpha \beta}{\alpha \beta}\right) x \\ & \Rightarrow x^2-\left(\frac{-\frac{b}{a}-2 \frac{c}{a}}{\frac{c}{a}}\right) x+\frac{1-\left(-\frac{b}{a}\right)+\frac{c}{a}}{\frac{c}{a}}=0\end{aligned}$
$\Rightarrow x^2-\left(\frac{-\frac{b}{a}-2 \frac{c}{a}}{\frac{c}{a}}\right) x+\frac{1-\left(-\frac{b}{a}\right)+\frac{c}{a}}{\frac{c}{a}}=0$
$\{$ from Eq. (i)\}
$\begin{aligned} & \Rightarrow x^2-\frac{(-b-2 c) x}{c}+\frac{a+b+c}{c}=0 \\ & \Rightarrow c x^2+(b+2 c) x+(a+b+c)=0\end{aligned}$
On comparing with $p x^2+q x+r=0$, we get $r=a+b+c$