Identify each of the following functions from $R^2$ to $R^2$ with their matrix representations : $f(x,y)=(x,0)$ , $g(x,y)=(2 x,0)$ , $h(x,y)=(0,y)$ , $i(x,y)=((x+y)/ 2,(x+y)/2)$. Now $f,g$ have the same eigenvectors but different eigenvalues. And $h,i$ have the same eigenvalues but different eigenvectors.
When finding common vectors in the column spaces of the matrices $A$ and $B$ (since in this case you only have two vectors(columns) per matrix you know they are not linearly dependent because they are not multiples of one another)
you are looking for common vectors in the spans of the form $a_1 \begin{bmatrix} 1\\1\\1\end{bmatrix}+a_2\begin{bmatrix} 2\\3\\2\end{bmatrix} = A\begin{bmatrix} a_1\\a_2\end{bmatrix} = A\vec{x} = b_1\begin{bmatrix} 5\\6\\5\end{bmatrix} + b_2\begin{bmatrix} 4\\3\\1\end{bmatrix} = B\begin{bmatrix} b_1\\b_2\end{bmatrix}=B\vec{y}$
(can a linear combination of the column vectors of $A$ equal a linear combination of the column vectors of $B$, i.e. the spaces share a common vector)
and $[\: A \:\: B \:] \begin{bmatrix} \vec{x}\\ -\vec{y}\end{bmatrix} = \vec{0}$ means essentially the same thing :=
$\begin{bmatrix} 1&2&5&4\\1&3&6&3\\1&2&5&1\end{bmatrix}\begin{bmatrix} a_1\\a_2\\-b_1\\-b_2\end{bmatrix} = a_1 \begin{bmatrix} 1\\1\\1\end{bmatrix}+a_2\begin{bmatrix} 2\\3\\2\end{bmatrix} - b_1\begin{bmatrix} 5\\6\\5\end{bmatrix} - b_2\begin{bmatrix} 4\\3\\1\end{bmatrix} = A\vec{x} -B\vec{y}=\vec{0}$
Since you are in $3$-space and you have two linearly independent vectors for each column space of $A$ and $B$, the subspaces generated by the span of those column vectors will be planes (if you had just one vector, the subspace would be a line, three linearly independent vectors, all of $3$-space), and as mentioned by the textbook answer, there are exists at least one nonzero solution (every homogeneous set of equations is consistent, has the zero solution) to the homogeneous set of equations
$$[\: A \:\: B \:] \begin{bmatrix} \vec{x}\\ -\vec{y}\end{bmatrix} = \vec{0} $$
because there are more unknowns than equations. So the planes either intersect in a line, or completely overlap.
This means the column spaces are not orthogonal by the definition of orthogonal because since they intersect, you can find a nonzero vector $\vec{v}$ s.t. $\vec{v}\in \text{col}(A),\text{col}(B)$ and $\vec{v} \cdot \vec{v} = ||\vec{v}||^2 \not= 0$.
Best Answer
As mentioned above, the cross product gives a formula and an immediate answer. However, if you're not familiar with the cross product, and as you've already hinted for the answer in your question, you can do it directly from the orthogonality condition.
Let $(a,b,c)$ be the vector, then the orthogonality reads
$$(a,b,c)\cdot (1,-1,1)=0\Leftrightarrow a-b+c=0,$$ $$(a,b,c)\cdot (1,0,-1)=0\Leftrightarrow a-c=0.$$
So $c=a$, and $b =2a$. We are free to choose any nonzero $a$, and letting $a=1$, we obtain $(1,2,1)$.