Answer
The limit is
$$\lim_{(x,y,z)\to(1,3,4)}\sqrt{x+y+z}=2\sqrt2.$$
The function is continuous everywhere in its' domain.
Work Step by Step
We will calculate the limit by simple substitution:
$$\lim_{(x,y,z)\to(1,3,4)}\sqrt{x+y+z}=\sqrt{1+3+4}=\sqrt{8}=2\sqrt2.$$
This function is continuous everywhere at its' domain because we can always perform the substitution and get
$$\lim_{(x,y,z)\to(x_0,y_0,z_0)}\sqrt{x+y+z}=\sqrt{x_0+y_0+z_0},$$
where the ordered triple $(x_0,y_0,z_0)$ is from the domain.