Let the total weight of the uniform beam be $W$ and its length be $L$. The center of gravity is at $L/2$ from either end. For the load to be shared equally, each person supports a force of $W/3$. The first person lifts one end (let's call it A, at position $x=0$) with an upward force $F_1 = W/3$. The other two people use a crossbar, so the total upward force from the crossbar is $F_{23} = W/3 + W/3 = 2W/3$. Let the position of the crossbar be $x$.

For the beam to be in rotational equilibrium, the net torque about any point must be zero. Taking torques about end A:

$$\sum \tau_A = F_{23} \cdot x - W \cdot \frac{L}{2} = 0$$ $$\left(\frac{2W}{3}\right)x = W\frac{L}{2}$$

Solving for $x$:

$$x = \frac{3}{2W} \cdot \frac{WL}{2} = \frac{3}{4}L$$

The crossbar should be placed at a distance of $3/4$ of the beam's length from the end held by the single person.