The statement generalizes to all dimensions.
Given a vector space $\mathbb{R}^n$ with the usual Euclidean metric, we represent a $k$-dimensional subplane spanned by the vectors $v_1,\dots,v_k$ with $v_1 \wedge v_2 \wedge \dots \wedge v_k$, where the wedge $\wedge$ is the antisymmetric linear product of the exterior algebra. Also, we have that the volume of an $n$-dimensional "parallelogramm" spanned by $v_1,\dots,v_n$ is just the $\lvert V \rvert$ in
$$ v_1 \wedge v_2 \wedge \dots \wedge v_n = V e_1 \wedge \dots \wedge e_n $$
where $e_i$ is the standard basis. Additionally, we have the Hodge star $\star$, which, in particular, sends a $n$-plane $v_1 \wedge \dots \wedge v_n$ to its volume $\mathrm{vol}(v_1 \wedge \dots \wedge v_n)$.The $n$-dimensional version of Lami's theorem is now:
Given $n+1$ vectors $v_i$ whose sum is zero, $$ \frac{\lvert v_i \rvert}{\mathrm{vol}(\hat{v}_{j_1} \wedge \dots \hat{v}_{j_n} )} = \frac{\lvert v_k\rvert}{\mathrm{vol}(\hat{v}_{l_1} \wedge \dots \wedge \hat{v}_{l_n})}$$
where the $j_{\circ}$ are a permutation of the indices except $i$, and likewise the $l_\circ$ a permutation of the indices except $k$.
Proof: Choose any two indices $i,j$, let them w.l.o.g. be $1,2$, and then apply $\wedge v_3 \wedge v_4 \wedge \dots \wedge v_{n+1}$ to the equation $\sum_i v_i = 0$. Antisymmetry of the wedge implies that a summand becomes zero as soon as one of the $v_i$ appears twice. Only the summands $v_1$ and $v_2$ survive this process, and we get
$$ v_1 \wedge v_3 \wedge \dots \wedge v_n = - v_2 \wedge v_3 \wedge \dots \wedge v_n $$
Now, write $v_i = \lvert v_i \rvert \hat{v}_i$ where $\hat{v_i}$ is a unit vector. Linearity of the wedge implies $v_1 \wedge v_3 \wedge \dots \wedge v_n = \lvert v_1 v_3 \dots v_n \rvert \hat{v}_1 \wedge \hat{v}_3 \wedge \dots \wedge \hat{v}_n$, and thus we arrive at
$$ \lvert v_1 \rvert \hat{v}_1 \wedge \hat{v}_3 \wedge \dots \wedge \hat{v}_n = \lvert v_2 \rvert \hat{v}_2 \wedge \hat{v}_3 \wedge \dots \wedge \hat{v}_n$$
and applying the Hodge star to both sides yields the desired result.
If I instead choose another origin located R away from point P, the
net torque is $$(\mathbf r_1+\mathbf R) \times \mathbf F_1 + (\mathbf r_2+\mathbf R) \times \mathbf F_2 + (\mathbf r_3+\mathbf R) \times \mathbf F_3 = \mathbf R \times (\mathbf F_1+\mathbf F_2+\mathbf F_3)$$
is apparently not zero.
While we often assume a net torque will give us rotation, really a net torque means the angular momentum about that point is changing, and that doesn't always imply a rotation.
If your new point $P$ is fixed on the object (but not at the center of mass), then the non-zero net force means that your choice of axis is accelerating and is not at rest in an inertial frame.
If your new point $P$ is at rest, then the object will accelerate from it. If the velocity vector for the center of mass does not pass through this new axis, then the angular momentum of the object about that axis is changing. The non-zero net torque you calculate shows the change in angular momentum, not necessarily a rotation.
If you want to determine that an object is in rotational equilibrium, you want to find a net torque of zero and one additional constraint, either
- The axis coincides with the center of mass or
- The object and the axis are not accelerating
Best Answer
Your basis vectors are unit vectors that are (ironically) unitless. So for all of the examples of basis vectors you propose, specifying what the vector represents is somewhat misleading.
In other words, you can think of all of your basis vectors ($\hat d_1$, $\hat d_2$, $\hat d_3$, $\hat F_1$, $\hat F_2$, $\hat F_3$, $\hat i$, $\hat j$, $\hat k$) as just directions in space. These sets of basis vectors do not represent positions, forces, etc. Your actual component magnitudes ($a$, $b$, $c$, etc.) have units, and this is what determines what the vector physically represents.
This is why you can explain many vector quantities with the same set of basis vectors. Position, force, etc. all have some direction in the space you are looking at. Therefore, you can express each vector using the same basis vectors.
So, when you say, "I can find three displacement vectors and use them as basis vectors," what you really mean is, "I have three displacement vectors, and I can take their directions and define basis vectors." (assuming they actually form a valid basis). With your example, these are $$\hat d_1=\frac{\mathbf d_1}{|\mathbf d_1|}$$ $$\hat d_2=\frac{\mathbf d_2}{|\mathbf d_2|}$$ $$\hat d_3=\frac{\mathbf d_3}{|\mathbf d_3|}$$
These basis vectors are just directions, they have no units associated with them, and you can use them to explain any other vector. For example, a force could be $$\mathbf F=F_1\,\hat d_1 +F_2\,\hat d_2+F_3\,\hat d_3$$ In other words, you have broken your force vector into components where each component is along the direction of each of displacement vectors you used to define your basis. The values $F_i$ have units of force, and this is what makes the vector a force vector. Also, note, for example, that the term $F_1\,\hat d_1$ is not "changing the intensity" of $\hat d_1$. It is just a scalar multiplication of the unit vector $\hat d_1$. This is analogous to how computing $3\cdot 4=12$ does not change what $4$ actually is.