You cannot say much about the relation between the two spans, except that they have the same dimension (for instance if one is the whole space, the other must be so as well). One way to see that you cannot expect a strict relation is to change the vectors in a way that certainly leaves their span unchanged, for instance permuting them or adding a multiple of one to another. This modifies the "transposed" vectors in a way that is not guaranteed to leave their span unchanged. To simplify your example, start with two vectors equal to zero.
Now that the question says that the original vectors $\vec a,\vec b,\vec c$ do indeed span the space, let me hint at an argument that the transposed ones do so as well. The vectors $\vec a,\vec b,\vec c$ form a basis of $\Bbb R^3$. Thus the map $\phi:\Bbb R^3\to\Bbb R^3$ that sends $(x,y,z)$ to the linear combination $x\vec a+y\vec b+z\vec c$ is an isomorphism (invertible linear map). If you interpret the transposed vectors as $1\times 3$ matrices $(a_i~b_i~c_i)$ (that is, you undo the transposition) then they describe the linear maps $\Bbb R^3\to\Bbb R$ defined by applying $\phi$ and then taking coordinate$~i$ (for $i=1,2,3$, repectively). Given this, can you argue that any $1\times 3$ matrix must be a linear combination of these three matrices, or equivalently that any linear map $\Bbb R^3\to\Bbb R$must be a linear combination of these three linear maps? (This is certainly true for the three coordinate maps themselves.)