Suppose we have $$ \begin{bmatrix} a & 2c & 2b \\ b & a & 2c \\ c & b & a\end{bmatrix} \times \begin{bmatrix} e \\ f \\g \end{bmatrix} = \begin{bmatrix} B+C+D \\ A+C+E \\A+B+F \end{bmatrix}$$ Where $$\begin{equation*} \begin{aligned} A &= b\big(e+f\big) \\ B &= a\big(e+g\big) \\ C &= 2c\big(f+g\big) \\ D &= g\big(-2(c-b)-a\big) \\ E &= f\big(a-(b+2c)\big) \\ F &= e\big((c-b)-a\big) \end{aligned} \end{equation*}$$
Is there a better solution?
Any improvement or suggestion will be highly appreciated. I mean to decrease the number of operations that are multiplication, addition and/or subtraction.
You have six unknowns and three equations, so you are free to choose any value for three of your unknowns. It is "nicest" to choose the following:
$A=B=C=0$,
$D = ae + 2cf + 2bg$,
$E = be + af + 2cg$,
$F = ce + bf + ag$.
To convince you that anything else would make it worse, let's look at the first two equations in your system,
$ae + 2cf + 2bg = B + C + D$,
$be + af + 2cg = A + C + E$.
Imagine choosing $C$ to be anything other than $0$, in fact, let's do as you chose above, $C=2cf + 2cg$. But the left side of the first equation does not contain $2cg$, and you had to account for this by giving to $D$ the extra summand $-2cg$. Similarly, the left side of the second equation does not have $2cf$, and you had to account for this by giving to $E$ the extra summand $-2cf$.
Anything you give to $A$, $B$, or $C$ has to be counteracted by one of your other three variables, so your simplest solution should have these three as $0$.