3. Given
$$
5 x+16 y+33 z \geqslant 136\left(x 、 y 、 z \in \mathbf{R}_{+}\right) \text {. }
$$

then the minimum value of $x^{3}+y^{3}+z^{3}+x^{2}+y^{2}+z^{2}$ is
3. 50 .

Notice that,
$$
\begin{array}{l}
x^{3}+x^{2}-5 x+3=(x+3)(x-1)^{2} \geqslant 0 \\
\Rightarrow x^{3}+x^{2} \geqslant 5 x-3, \\
y^{3}+y^{2}-16 y+20=(y+5)(y-2)^{2} \geqslant 0 \\
\Rightarrow y^{3}+y^{2} \geqslant 16 y-20, \\
z^{3}+z^{2}-33 z+63=(z+7)(z-3)^{2} \geqslant 0 \\
\Rightarrow z^{3}+z^{2} \geqslant 33 z-63 .
\end{array}
$$

Therefore, $x^{3}+y^{3}+z^{3}+x^{2}+y^{2}+z^{2}$
$$
\geqslant 5 x+16 y+33 z-86=50 \text {. }
$$

The minimum value 50 is achieved when and only when $x=1, y=2, z=3$.