Why are these vectors expressed as row vectors and not column vectors? When to write as row vectors or column vectors? 

Everytime I have been asked to find a basis when the vectors were given in comma delimited form, I, and the book, would write out the vectors as columns in a matrix.  Another example in the book asks a similar question, "find the null space spanned the vectors...", and they have expressed the vectors given in comma delimited form as columns in a matrix.  I tried doing this question over with the vectors expressed as row vectors in the matrix but it does not give me the same answer.  Why is this example expressing the vectors as row vectors in the matrix?  And why does it not work when I write out these vectors as columns in a matrix?

 A: You should indeed get the same answer if you express the row vectors $w_1,w_2,w_3,w_4$ as columns in a matrix, as long as you set the problem up in the correct manner.
The solution, as it is given in the text that you quote, expresses the row vectors $w$ as rows in a matrix $A$ and then asks you to solve for the unknown column vector $V$ in the equation $AV=0$. You get a solution with basis column vectors $v_1,v_2,v_3$ as written, and the entire solution could also be expressed parametrically as
$$x_1 = -3r - 4s - 2t$$
$$x_2 = 1r + 0s + 0t$$
$$x_3 = 0r - 2s + 0t$$
$$\vdots$$
or more briefly as $x=r\, v_1 + s\, v_2 + t \,v_3$.
Now let's do it using columns. First, convert the $w$'s into column vectors by taking their transposes: $w_1^T, w_2^T, w_3^T, w_4^T$. Next, collect those column vectors into a matrix, which is simply the transpose of the original matrix, namely $A^T$. Next, convert the column vector $V$ into a row vector by taking its transpose: $V^T = (x_1, x_2, x_3, x_4, x_5, x_6)$. Finally, write out the matrix equation that you must solve for the $x$'s:
$$V^T A^T = 0
$$
Note the different order of $V$ and $A$. This equation is equivalent to
$$A V = 0
$$
because $(AV)^T = V^T A^T$. So you will get the same parametric solutions for the $x$'s, and the null space will have the same basis vectors, except that they have been transposed into the row vectors $v_1^T, v_2^T, v_3^T$.
Edit Here are some additional remarks addressing the comment. 
Row reduction of $A$ corresponds to multiplying $A$ on the left by a square elementary matrix $M$: the effect of the row reduction is to replace $A$ by $MA$. This does not alter the "null space" which is the solution set of the equation $AV=0$, because $AV=0$ if and only if $MAV=0$ (since $M$ is invertible).
Similarly, column reduction of $A^T$ corresponds to multiplying $A^T$ on the right by a square elementary matrix $N$, and this does not alter the solution set of the equation $V^T A^T=0$ because $V^T A^T = 0$ if and only if $V^T A^T N = 0$ (again since $N$ is invertible).
On the other hand, column reduction on $A$ may indeed alter the solution set of $AV=0$, because $AV=0$ and $ANV=0$ need not have the same solution sets. Similarly, row reduction on $A^T$ does indeed alter the solution set of $V^T A^T = 0$, because $V^T A^T = 0$ and $V^T M A^T$ need not have the same solution set.
The common theme here is that matrix multiplication is not commutative; multiplying on the left and on the right have different effects; row operations and column operations have different effects.
A: If you are trying to find vectors orthogonal to a given set of vectors, it just feels more comfortable to view them as rows of matrix so you can solve for the null space of the matrix to get the orthogonal complement to the given set. The real lesson here, is that when you are done here, the orthogonal complement of the given vectors must be of the same form as the given vectors, in this case, back as rows.
The problem could have started with column vectors where you "flip" them to rows solve for the null space (as column vectors) and be done. 
