TITLE: Symmetric determinant
QUESTION [5 upvotes]: So, basically i want to prove that the value of $$\begin{vmatrix}
ax-by-cz & ay+bx & cx+az\\
ay+bx & by-cz-ax & bz+cy\\
cx+az & bz+cy & cz-ax-by\\
\end{vmatrix}$$
is equal to $$(x^2+y^2+z^2)(a^2+b^2+c^2)(ax+by+cz)$$
$\mathbf {What}$ $\mathbf {I} $ $\mathbf{have} $ $\mathbf{tried}$
: I have noticed it is a symmetric matrix but cannot proceed on that thought further. Next, I have tried multiplying row $1,2,3$ with $yz, xz, xy$ respectively with no luck. I also did some transformations but they were of no use again.

Thanks in advance.

REPLY [4 votes]: Write column vectors $A,X$ as evident. Your matrix is the result of beginning with
$$ A X^T + X A^T   $$ and subtracting (using traditional dot product)
$$  (A \cdot X) I.  $$
The usefulness is that we know the eigenpairs (eigenvalue,eigenvector) of $A X^T + X A^T,$ namely (using traditional cross product)
$$ 0, \; \; A  \times X,  $$
$$ (A \cdot X) + |A||X|, \; \; |X|A + |A|X, $$
$$ (A \cdot X) - |A||X|, \; \; |X|A - |A|X. $$
Now subtract off $(A \cdot X)I,$ the eigenpairs are
$$ -(A \cdot X), \; \; A  \times X,  $$
$$   |A||X|, \; \; |X|A + |A|X, $$
$$  - |A||X|, \; \; |X|A - |A|X. $$
The product of the eigenvalues is
$$ (A \cdot X) |A|^2 |X|^2  $$
EXAMPLE:
$$ A^T = (2,3,6), \; \; |A| = 7, $$
$$ X^T = (2,6,9), \; \; |X| = 11, $$
$$ A \cdot X = 76 $$
as they are close to parallel.
$$
A X^T + X A^T =
\left(
\begin{array}{rrr}
8 & 18 & 30 \\
18 & 36 & 63 \\
30 & 63 & 108
\end{array}
\right)
$$
subtract $76I,$
$$
\left(
\begin{array}{rrr}
-68 & 18 & 30 \\
18 & -40 & 63 \\
30 & 63 & 32
\end{array}
\right)
$$
The eigenvectors are the cross product (divided out a 3), then $11A + 7 X,$ which I divided by a common 3, then $11A - 7 X.$
Matrix with eigenvectors as columns 
$$
\left(
\begin{array}{rrr}
3 & 12 & 8 \\
2 & 25 & -9 \\
-2 & 43 & 3
\end{array}
\right)
$$
eigenvalues
$$ -76, 77, -77  $$