Write $1_A$ for the <indicator function> and use the normalized <Fourier analysis on a finite abelian group>
$$
\widehat A(r)=\mathbb E_x1_A(x)\omega^{-rx}.
$$
There are $N^5$ sextuples satisfying the equation, since any five coordinates determine the sixth. The desired <probability> is consequently
$$
\frac1{N^5}\sum_{x_1+ x_2+x_3=x_4+x_5+x_6}\prod_{j=1}^6 1_A(x_j).
$$
By <orthogonality of complex exponentials>, the indicator of the equation is
$$
\frac1N\sum_{r\in\mathbb Z_N}
\omega^{r(x_1+x_2+x_3-x_4-x_5-x_6)}.
$$
Substitution makes all six sums independent and gives the <sixth Fourier moment as a three-sum collision count>:
$$
\sum_{r\in\mathbb Z_N}\widehat A(-r)^3\widehat A(r)^3.
$$
Because $1_A$ is real, $\widehat A(-r)=\overline{\widehat A(r)}$. Hence the probability is
$$
\boxed{\sum_{r\in\mathbb Z_N}|\widehat A(r)|^6.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 147 / 1 / ii / Solution

For $A$ in a finite <abelian group> and normalized <Fourier transform> $\widehat{1_A}$, the normalized number of sextuples satisfying
$$
x_1+x_2+x_3=x_4+x_5+x_6
$$
is
$$
\sum_{\gamma\in\widehat G}|\widehat{1_A}(\gamma)|^6.
$$
This follows by inserting <orthogonality of complex exponentials> for the displayed equation: the three variables on one side contribute $\widehat{1_A}(\gamma)^3$ and those on the other side contribute its <complex conjugate>.

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