If $$x + y + z = 3, xy + yz + zx = -12$$ and $$xyz = -16$$, then the value of $$\sqrt{x^3 + y^3 + z^3 + 13}$$ is:

$$x+y+z=3$$

$$x+y=3-z$$........(1)

$$\left(x+y\right)^3=\left(3-z\right)^3$$

$$x^3+y^3+3xy\left(x+y\right)=27-z^3-3.3.z\left(3-z\right)$$

$$x^3+y^3+3xy\left(3-z\right)=27-z^3-9z\left(x+y\right)$$  [From (1)]

$$x^3+y^3+9xy-3xyz=27-z^3-9xz-9yz$$

$$x^3+y^3+z^3=27-9xy-9xz-9yz+3xyz$$

$$x^3+y^3+z^3=27-9\left(xy+yz+zx\right)+3xyz$$

$$x^3+y^3+z^3=27-9\left(-12\right)+3\left(-16\right)$$

$$x^3+y^3+z^3=27+108-48$$

$$x^3+y^3+z^3=87$$.......(2)

$$\sqrt{x^3+y^3+z^3+13}=\sqrt{87+13}$$

$$=\sqrt{100}$$

$$=10$$

Hence, the correct answer is Option C