Prove that

Let us assume that $\frac{(3-4 \sqrt{2})}{7}$ is a rational number.

Thus, $\frac{(3-4 \sqrt{2})}{7}$ can be represented in the form of $\frac{p}{q}$, where $p$ and $q$ are integers, $q \neq 0, p$ and $q$ are co-prime numbers.

$\frac{3-4 \sqrt{2}}{7}=\frac{p}{q}$

$\Rightarrow 3-4 \sqrt{2}=\frac{7 p}{q}$

$\Rightarrow 4 \sqrt{2}=3-\frac{7 p}{q}$

$\Rightarrow 4 \sqrt{2}=\frac{3 q-7 p}{q}$

$\Rightarrow \sqrt{2}=\frac{3 q-7 p}{4 q}$

Since, $\frac{3 q-7 p}{4 q}$ is rational $\Rightarrow \sqrt{2}$ is rational

But, it is given that $\sqrt{2}$ is an irrational number.

Therefore, our assumption is wrong.

Hence, $\frac{3-4 \sqrt{2}}{7}$ is an irrational number.