For simplicity's sake, I'm working with a 3×3 square matrix in which none of the column or row vectors is the zero vector.
I tried graphing the columns of the matrix {{1,4,-3},{2,7,-5},{3,6,-3}} (where each group of $3$ is a row) and those $3$ vectors are on the same plane. I understand why that's the case; what I don't understand is why it necessarily implies that the $3$ row vectors are also on a single plane.
I have some algebraic intuition for the 2d case. For example with vectors v = <1,2> and w = <3,6>, the fact that w = 3v means that the respective components of the two vectors are in the same proportion, and therefore the vectors x = <1,3> and y = <2,6> are also colinear.
I tried to extend that kind of logic to the 3d case but I didn't manage to because of the increased complexity. For ex in the 3×3 matrix that I used, if column vectors are u, v and w, then u = v + w. Since one vector isn't simply a scalar multiple of another as in the 2d case, I can't seem to apply the same logic of proportions kept.
I saw this question: For a square matrix, row vectors are linearly independent if and only if columns are., but it doesn't give the kind of intuition I'm looking for.
Best Answer
If the rows are coplanar, then there exists a nonzero linear combination $x_1R_1+x_2R_2+x_3R_3=0$ with at least one of $x_1,x_2,x_3\ne 0$. But this says that $LM=0$, where $L$ is the row vector $L=(x_1,x_2,x_3)$ and $M$ is the matrix with rows $R_1,R_2,R_3$. Then $Me_i$ is the $i$th column, $C_i$, of the matrix where $e_i$ is the $i$th standard basis vector, so $LMe_i=0$ for all $i$. Hence $LC_i=0$ for all $i$. Thus all the columns of $M$ lie in the plane $x_1x+x_2y+x_3z=0$.
I've added an answer with a more general explanation of how this works to the question Will Jagy linked in the comments, though I'm not sure if the general case will be useful to you or not.