The answer is correct. For this aim, you can consider
the following consequence:
$$f_1=r_1^{-1}\cdot z_1,~f_2=r_2^{-1}\cdot z_2,~f_3=r_3^{-1}\cdot z_3$$
Then we have from the third component $u_1$ and $u_2$:
$$(r_1+r_2-r_3)\cdot d=\alpha$$
$$(z_1+z_2-z_3)\cdot d = (r_1 \cdot f_1+r_2 \cdot f_2-r_3 \cdot f_3)\cdot d=\beta$$
Using the multiplication $f_1$, $f_2$ and $f_3$ in the above formula (first) and subtracting from the second, we obtain the following three equations:
$$ r_2 \cdot d \cdot (f_1-f_2) - r_3 \cdot d \cdot (f_1 - f_3)= \alpha \cdot f_1 - \beta$$
$$ r_1 \cdot d \cdot (f_2-f_1) - r_3 \cdot d \cdot (f_2 - f_3)= \alpha \cdot f_2- \beta $$
$$ r_1 \cdot d \cdot (f_3-f_1) - r_2 \cdot d \cdot (f_3 - f_2)= \alpha \cdot f_3 - \beta$$
In the next step, let $x_1=r_1 \cdot d, x_2=r_2.d, x_3=r_3 \cdot d$, thus we have:
$$ x_2 \cdot (f_1-f_2) - x_3 \cdot (f_1 - f_3)= \alpha \cdot f_1 - \beta$$
$$ x_1 \cdot (f_2-f_1) - x_3 \cdot (f_2 - f_3)= \alpha \cdot f_2 - \beta$$
$$ x_1 \cdot (f_3-f_1) - x_2 \cdot (f_3 - f_2)= \alpha \cdot f_3 - \beta$$
Finally, we solve the linear equation system. Therefore, we obtain values $x_1$, $x_2$ and $x_3$.