Linear program in three variables $$\begin{array}{ll} \text{minimize} & 2x - 3y - z\\ \text{subject to} & x + y + z = 1\\ & x,y,z \geq 0\end{array}$$
One possible idea that comes into mind is the Lagrange multiplier approach where we could let $f(x,y,z) = 2x-3y-z, \quad $ $g(x,y,z) = x+y+z=1$ and $h(x,y,z) = x,y,z$; so that we could use
$$\frac{\partial f}{\partial x} = \lambda\frac{\partial g}{\partial x} + \mu\frac{\partial h}{\partial x}, \quad \frac{\partial f}{\partial y} = \lambda\frac{\partial g}{\partial y} + \mu\frac{\partial h}{\partial y} \text{ and } \frac{\partial f}{\partial z} = \lambda\frac{\partial g}{\partial z} + \mu\frac{\partial h}{\partial z}$$
But after second thought, I realized that $h(x,y,z)$  may not be a function which makes it difficult to use the Lagrange multiplier formula. I know there are multiplies way of solving such minimization problem (including graphical method in 3D - which is not readily available online, most online calculators are 2D) but I really don't know which one is the most convenient for this problem. Would appreciate any explicit solution to the problem.
 A: The linear programming dual problem is to maximize $\lambda$ subject to:
\begin{align}
\lambda &\le 2\\
\lambda &\le -3\\
\lambda &\le -1
\end{align}
Obviously, $\lambda = -3$ is optimal.  Complementary slackness then implies that $x=0$ and $z=0$, so the primal equality constraint implies that $y=1$.
A: The much easier one is to solve for $z = 1-x-y$ and minimize $$2x-3y-(1-x-y) = 3x-2y-1$$ over $(x,y) \in [0,1]^2$.

UPDATE
Since you are minimizing linear function over $[0,1]$, best $x$ is at $1$ and $y = 0$ (since coefficient of $x$ is positive and coefficient of $y$ is negative). Hence, $x=1,y=0$ and this implies $z=0$ and you are done.

UPDATE 2
You are originally optimizing with constraints $x,y,z \ge 0$ and $x+y+z = 1$. The first constraint implies $x,y \ge 0$, and the second implies $x,y \le 1$, so you must have $0 \le x,y \le 1 \iff (x,y) \in [0,1]^2$.
As for producing minimum value, you are minimizing $F(x,y)=3x-2y-1$, and since $x$ has a positive coefficient, you have to pick the smallest possible value for $x$. Since $y$ has a negative coefficient in $F$, you want to pick the largest possible value for $y$, so the minimum occurs at $x=0,y=1$ with $F(0,1) = -3$.
If you were maximizing, the logic would flip and you would get $x=1,y=0$ with $F(1,0) = 2$.
A: This is a linear programming problem with compact feasible space, hence the optimal solution exists.
An optimal solution exists at the corner.
We want $y$ to be as large as possible. The optimal solution is $(0,1,0)$.

If you want to do it by Lagrangian approach. You can consider $h_1(x,y,z) =x$, $h_2(x,y,z) =y$, $h_3(x,y,z)=z$ and assign Lagrange multiplier $\mu_i, i = 1,2,3$.
Edit: Your $\mu$ is not correct.
We have $$2-\lambda - \mu_1 = 0$$
$$-3-\lambda-\mu_2 = 0$$
$$-1-\lambda - \mu_3$$
where $\mu_i \ge 0$.
Hence, $$\mu_1 = 2-\lambda \ge 0$$
$$\mu_2 = -3-\lambda \ge 0$$
$$\mu_3 = -1 -\lambda \ge 0$$
Hence $-3 \ge \lambda$, and hence $\mu_1 > 0$ and $\mu_3 > 0$. Hence by complementary slackness condition, $x=z=0$ and we conclude that $y=1$.

Remark: This question has a physical meaning, you have a choice of distributing resources into $x, y, z$ where $y$ gives you the most benefit. The optimal strangety is to invest everything to $y$.
A: Hint :
the solution is simple : it is $x=0,y=1,z=0$, it is not difficult to see why, try to get why intuitively and then find a formal elementary method to prove it.
Method :
Say $2x-3y-z = f$.
Now, choose a $x = \epsilon>0$ but then $f = 2\epsilon-3y-z > -3y-z \quad\forall y,z \geq 0$. So you have to choose $x=0$.
Now you know that $z=1-y$ so $f =-2y-1$. So $\min(f) = -3$ when $y=1$ and $z=0$.
