Let's calculate the value of the expression $K$ when $x, y$, and $z$ are replaced by the first, second, and third numbers of the I-V. number triples, respectively. Then, let's factorize $K$ and explain the observations from the previous calculations.

$$
\begin{aligned}
K & =(x+2 y+3 z)(2 x-y-z)(y+2 z+3 x)+ \\
& +(y+2 z+3 x)(2 y-z-x)(z+2 x+3 y)+ \\
& +(z+2 x+3 y)(2 z-x-y)(x+2 y+3 z)
\end{aligned}
$$

I. $1,4,9 ; \quad$ II. $11,14,19 ; \quad$ III. $4,9,1$; IV. $1,8,11 ; \quad$ V. $5,8,-2$.
ג) Substitution leads to a surprising result, in the case of the I-III. number triples, we always get -286, while in the IV. and V. cases, we get 884 uniformly.

$\beta)$ Let us notice that the first factor of the second product of $K$ is identical to the third factor of the first product, the first factor of the 3rd product is the third factor of the 2nd product, and the third factor is the first factor of the 1st product, - furthermore, that in each product, the third factor is equal to the sum of the two preceding factors. We will use these in the desired transformation of the product. Let us denote the 1st, 2nd, and the 2nd factor of the 2nd product of the 1st product with a letter each:

$$
x+2 y+3 z=A, \quad 2 x-y-z=B, \quad 2 y-z-x=C
$$

then $K$ can be written, and transformed as follows:

$$
\begin{aligned}
K & =A B(A+B)+(A+B) \cdot C \cdot(A+B+C)+(A+B+C) \cdot[A-(A++C)] \cdot A= \\
& =A B(A+B)+(A+B+C)(B C-A B)=B\left[\left(A^{2}+A B\right)+\left(C^{2}-A^{2}\right)+(B C-\right. \\
& -A B)]=B C(B+C)=(-B) \cdot(-C) \cdot(B+C) .
\end{aligned}
$$

Here the three factors are the second factor of the first, second, and third products of $K$ multiplied by $(-1)$, therefore

$$
K=(y+z-2 x) \cdot(z+x-2 y) \cdot(x+y-2 z)
$$

ү) Here the second factor can be formed from the first, the third factor from the second (and finally the first from the third) by replacing the letters $x, y, z$ with $y, z, x$ respectively, in the usual expression: $x, y, z$ are cyclically replaced. Similarly, the III. number triple can be derived from the I., which explains why we got the same value for $K$ in the III. and I. cases, since in (1) form of $K$ only the order of the factors changes.

We also see that by substituting $x+u, y+u, z+u$ for $x, y, z$ respectively, where $u$ is an arbitrary number, the factors of (1) remain unchanged, because, for example, in the first one, $2 x$ changes by the same amount as $y+z$. This explains the equality obtained in the II. and I. number triples, $u=10$.

The explanation for the equality of the IV. and V. results uses both properties observed in (1): Let $u=-3$, then according to the second property, the IV. number triple gives the same value as the triple $x=1-3=-2$, $y=8-3=5, z=11-3=8$, from which the V. number triple can be derived by cyclic replacement.

Remark. $K$ can be simplified in many other ways. For example, by introducing a new letter for the first factor of the products:

$$
x+2 y+3 z=D, \quad y+2 z+3 x=E, \quad z+2 x+3 y=F
$$

we can consider our equations as a system of equations, with $x, y, z$ as unknowns and $D, E, F$ as parameters. From this,

$$
x=\frac{1}{18}(-5 D+7 E+F), \quad y=\frac{1}{18}(D-5 E+7 F), \quad z=\frac{1}{18}(7 D+E-5 F)
$$

with these, the second factor of the 1st product is

$$
2 x-y-z=\frac{1}{18}(-18 D+18 E)=E-D
$$

which we also noticed above. Similarly, the second factor of the 2nd and 3rd products is $F-E, D-F$. The form of $K$ thus obtained

$$
S=D \cdot(E-D) \cdot E+E \cdot(F-E) \cdot F+F \cdot(D-F) \cdot D
$$

is not difficult to bring to the form $(E-D)(F-E)(D-F)$.

However, a similar transformation could not be performed, for example, by taking the factors of the 1st product as parameters, because - as we have seen - these are not independent of each other.