We have, $\alpha, \rho$ and $\gamma$ are roots of equation
$$
\begin{gathered}
x^{3}+p x+q=0 \\
\alpha+\beta+\gamma=0 \\
\text { Here, } \alpha \beta+\beta \gamma+\gamma \alpha=p \\
\alpha \beta \gamma=-q \\
\text { Applying } C_{1} \rightarrow C_{1}+C_{2}+C_{3} \text {, we get } \\
\left|\begin{array}{lll}
\alpha+\beta+\lambda & \beta & \gamma \\
\alpha+\beta+\lambda & \gamma & \alpha \\
\alpha+\beta+\lambda & \alpha & \beta
\end{array}\right|=(\alpha+\beta+\lambda)\left|\begin{array}{ccc}
1 & \beta & \gamma \\
1 & \gamma & \alpha \\
1 & \alpha & \beta
\end{array}\right|
\end{gathered}
$$
Applying $R_{2} \rightarrow R_{2}-R_{1}, R_{3} \rightarrow R_{3}-R_{1}$, we get
$$
\begin{aligned}
&(\alpha+\beta+\lambda)\left|\begin{array}{ccc}
1 & \beta & \gamma \\
0 & \gamma-\beta & \alpha-\gamma \\
0 & \alpha-\beta & \beta-\gamma
\end{array}\right| \\
&=(\alpha+\beta+\lambda)[(\gamma-\beta)(\beta-\gamma)-(\alpha-\gamma)(\alpha-\beta)] \\
&=0 \times[(\gamma-\beta)(\beta-\gamma)-(\alpha-\gamma)(\alpha-\beta)]=0
\end{aligned}
$$