we have $x+y+z=12$...(i)
$$
\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2=50...(ii)
$$
and $\mathrm{x}^3+\mathrm{y}^3+\mathrm{z}^3=216$...(iii)
$$
\begin{aligned}
& \because(\mathrm{x}+\mathrm{y}+\mathrm{z})^2-\left(\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2\right)=2(\mathrm{xy}+\mathrm{yz}+\mathrm{zx}) \\
& \Rightarrow \mathrm{xy}+\mathrm{yz}+\mathrm{zx}=47
\end{aligned}
$$
now $x^3+y^3+z^3-3 x y z=(x+y+z)\left(x^2+y^2+z^2-x y-y z-z x\right)$
$$
\Rightarrow \mathrm{xyz}=60
$$
now from equn (iii) $\mathrm{y}^3+\mathrm{z}^3=216-\mathrm{x} 3$
$$
\Rightarrow(12-\mathrm{x})\left\{\left(50-\mathrm{x}^2\right)-\left[47-12 \mathrm{x}+\mathrm{x}^2\right]\right\}=216-\mathrm{x}^3
$$
on solving we get
$$
x^3-12 x^2+47 x-60=0...(iv)
$$
by hit at trial
$$
\begin{aligned}
& x=3 \text { is a root of equation IV } \\
& \Rightarrow(x-3)\left(x^2-9 x+20\right)=0 \\
& \Rightarrow(x-3)(x-4)(x-5)=0 \\
& \Rightarrow x=3,4,5
\end{aligned}
$$
now no of solution are $3 !=6$