It seems that the minimal value of $s=x+y+z$ is $a\sqrt{3}$, attained at the centroid. The maximal value of $s$ is $2a$, attained at a vertex.
By continuity of $s$, all intermediate values between $a\sqrt{3}$ and $2a$ are possible too.
The plane $\pi$ through $(1,1,0),(3,3,1),(6,1,0)$ has equation $y-2z=1$, so it is perpendicular to the vector $(0,1,-2)$. There is an only value of $\lambda$ for which:
$$ (9,5,0)+\lambda (0,1,-2) \in \pi $$
and that gives the projection of $P=(9,5,0)$ on $\pi$, that is the point on $\pi$ with the minimal distance from $P$. If we solve the previous equation, we have that the projection of $P$ on $\pi$ lies in:
$$ Q = \left(9,\frac{21}{5},\frac{8}{5}\right)=-\frac{39}{25}(1,1,0)+\frac{8}{5}(3,3,1)+\frac{24}{25}(6,1,0) $$
i.e. outside the triangle made by $A=(1,1,0),B=(3,3,1),C=(6,1,0)$. This gives that the solution is not inside $ABC$, so it is somewhere on the perimeter of $ABC$, maybe in a vertex. Among $A,B,C$, the closest point to $Q$ is $C$. Let we compute the squared distance of a point of $BC$ from $P$:
$$ \|t(3,3,1)+(1-t)(6,1,0)-(9,5,0)\|_{2}^{2} = 25+2t+14t^2 $$
The RHS is obviously an increasing function on $[0,1]$, hence the optimal point on the $BC$ edge is the vertex $C$. The same happens for the $AC$ edge, while on the $AB$ edge the winner is the $B$ vertex, so the solution is given by the vertex $\color{red}{C}$, with distance $\color{red}{5}$ from $P$.
Best Answer
We consider the function $$ f(x,y) = \sum_{j=1}^3 [(x-a_j)^2+(y-b_j)^2] $$ where $(a_j,b_j)$ the vertices of the triangle.
There is only one critical point that is $(\frac{a_1+a_2+a_3}{3},\frac{b_1+b_2+b_3}{3})$ which is a minimum and the centroid point of the triangle.
Hence, the point lied inside or on the triangle with the maximum total distance from the 3 vertices should lie on the boundary of the domain for $(x,y)$ i.e. on the triangle. This follows from the fact that $f: K\rightarrow \mathbb{R}$ where $K$ is the triangle with its interior, is a continuous function on a compact set so it attains a maximum value. Since it doesn't have a local maximum the maximum value is attained at the boundary.
Then as already suggested in another answer posted: let $A,B,C$ the vertices and assume that the point $D=(x,y)$ we are looking for, lies on $AB$. Then the total distance from the 3 vertices, let $d$, is equal to $d=AB+DC$. Moreover, $DC$ is bounded above by the $\max{\{AC,BC\}}$. Hence, $D$ is the vertex $A$ or $B$ if $AC\geq BC$ or $BC\geq AC$ respectively.
In other words the point that maximizes the distance from the 3 vertices of the triangle is the vertex whose adjacent edges are the two longest.