If $\alpha, \beta, \gamma$ be the zeroes of the polynomial $p(x)$ such that $(\alpha+\beta+\gamma)=3,(\alpha \beta+\beta \gamma+\gamma \alpha)$ $=-10$ and $\alpha \beta \gamma=-24$, then $\mathrm{p}(\mathrm{x})=?$ (a) $x^{3}+3 x^{2}-10 x+24$ (b) $x^{3}+3 x^{2}+10 x-24$ (c) $x^{3}-3 x^{2}-10 x+24$ (d) none of these

The correct option is option (c) $x^{3}-3 x^{2}-10 x+24$

$\alpha, \beta$ and $\gamma$ are the zeroes of polynomial $p(x)$.

$(\alpha+\beta+\gamma)=3,(\alpha \beta+\beta \gamma+\gamma \alpha)=-10$ and $\alpha \beta \gamma=-24$

$\therefore \mathrm{p}(\mathrm{x})=\mathrm{x}^{3}-(\alpha+\beta+\gamma) \mathrm{x}^{2}+(\alpha \beta+\beta \gamma+\gamma \alpha) \mathrm{x}-\alpha \beta \gamma$

$=x^{3}-3 x^{2}-10 x+24$