Symmetric determinant 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.
 A: 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
$$ -76, 77, -77  $$
A: If all else fails, use the formula or Rule of Sarrus for a $3 \times 3$ determinant, and expand.  It's messy, but...
A: Let $A=\begin{pmatrix}
ax-by-cz & ay+bx & cx+az\\
ay+bx & by-cz-ax & bz+cy\\
cx+az & bz+cy & cz-ax-by\\
\end{pmatrix}.$
Let 
$$\tag{1}B:=A+(ax+by+cz)I_3=\begin{pmatrix}
2ax & ay+bx & cx+az\\
ay+bx & 2by & bz+cy\\
cx+az & bz+cy & 2cz\\
\end{pmatrix}$$ 
and
$$\tag{2}V:=(bz-cy,cx-az,ay-bx)^T.$$
It is easy to show that $B*V=0$. Thus $\lambda_1:=-(ax+by+cz)$ is an eigenvalue of $A$.
An angle of attack is by using the fact that the determinant of a matrix is the product of its eigenvalues.  
Taking a look at the result, one should be tempted to consider that the other eigenvalues are $\pm(x^2+y^2+z^2)$ and $\pm(a^2+b^2+c^2)$. One cannot say such a thing, but, in fact, one can write by long division, the characteristic polynomial of matrix $A$ under the form : 
$$\chi_A(\lambda)=-(\lambda-\lambda_1)(\lambda^2-\mu) \ \ \text{with} \ \ \mu:=(x^2+y^2+z^2)(a^2+b^2+c^2).$$ 
proving that $\det(A)$ is the product $\lambda_1 * A$ as "minus the constant term of the characteristic polynomial". 
Remarks: 
1) Mathematica has been very useful...
2) I would bet that there is a geometrical interpretation for matrix $A$ ; because, for example, $ax+by+cz$ is the dot product of $(a,b,c)^T$ and $(x,y,z)^T$ whereas vector $V$ defined in (2) is their cross product. But I have been unable to find one...
Edit: the solution given by Will Jagy gives a very satisfying answer to Remark 2.
