Distance to a subspace

Orthogonal complement. $F^\perp = \operatorname{span}\bigl((1,1,1)\bigr)$, normalized to $u=(1,1,1)/\sqrt3$.

Projection onto $F^\perp$. $\langle v,u\rangle = (1+1+1)/\sqrt3 = 3/\sqrt3 = \sqrt3$. Hence
$$p_{F^\perp}(v) = \langle v,u\rangle\, u = \sqrt3\cdot\frac{(1,1,1)}{\sqrt3} = (1,1,1) = v.$$
This is consistent: $v\in F^\perp$ because $(1,1,1)$ is exactly the normal of $F$ (and $v\notin F$ since $1+1+1=3\neq 0$).

Distance. The distance to $F$ is $\|v-p_F(v)\|$. Since $v-p_F(v)\in F^\perp$ and $\mathbb{R}^3 = F\oplus F^\perp$, we have $v-p_F(v)=p_{F^\perp}(v)=v$. Therefore
$$\operatorname{dist}(v,F) = \|v\| = \sqrt{1^2+1^2+1^2}=\sqrt3.$$

Conclusion. The distance from $(1,1,1)$ to the plane $x+y+z=0$ is $\sqrt3$.