real analysis - How to show that $asqrt 6 + b sqrt 3 + c sqrt 2$ is irrational where $a,b,c in mathbb{Q}$

How can I show that $a\sqrt 6 + b \sqrt 3 + c \sqrt 2$ is irrational given that at least one of $b, c$ is nonzero? I've seen the argument when $a = b = c = 1$, but in the general case the algebra is a lot more tedious (I need to avoid division by $0$, for example). Here's what I have so far:

Suppose that $a\sqrt6 + b\sqrt3 + c\sqrt2 \in \mathbb{Q}$. If $a = 0$, then $b\sqrt3 + c\sqrt2 \in \mathbb{Q}$ (note that $b\sqrt3 + c\sqrt2 \neq 0$, otherwise $b\sqrt3 = -c\sqrt2 \implies \sqrt6 = -2c/b$ which is a contradiction since $\sqrt6$ is irrational) so that

$$\begin{align*} b\sqrt3 - c\sqrt2 = \frac{3b - 2c}{b\sqrt3 + c\sqrt2} \in \mathbb{Q} \end{align*}$$

Hence,

$$\begin{align*} (b\sqrt3 + c\sqrt2) + (b\sqrt3 - c\sqrt2) = 2b\sqrt3 \in \mathbb{Q} \end{align*}$$

which is a contradiction if $b \neq 0$. (Note that if $b = 0$, then by assumption $c \neq 0$ and we can use a similar argument to arrive at a contradiction)

Therefore, assume that $a \neq 0$.
Then, it follows that

$$\begin{equation*} r = (a\sqrt3 + c)\left(\sqrt2 + \frac{b}{a}\right) = (a\sqrt6 + b \sqrt3 + c\sqrt2) + \frac{cb}{a} \in \mathbb{Q} \end{equation*}$$

since the sum of two rationals is rational. Then,

$$\begin{align*} a\sqrt3 + c &= \frac{r}{\sqrt2 + \frac{b}{a}} \\&= \frac{r}{\sqrt2 + \frac{b}{a}} \cdot \frac{\sqrt2 - \frac{b}{a}}{\sqrt2 - \frac{b}{a}} \\&= \frac{r}{2 - \frac{b^2}{a^2}}\cdot\left(\sqrt2 - \frac{b}{a}\right) \end{align*}$$

and so

$$\begin{align*} a\sqrt3 - \frac{r\sqrt2}{2 - \frac{b^2}{a^2}}&= \frac{-rb}{a\left(2 - \frac{b^2}{a^2}\right)}-c \in \mathbb{Q} \end{align*}$$

Note that $a\sqrt3 - \frac{r\sqrt2}{2 - \frac{b^2}{a^2}} \neq 0$, otherwise $a\sqrt6 = \frac{2r}{2 - \frac{b^2}{a^2}} \in \mathbb{Q}$. On the other hand,

$$\begin{align*} a\sqrt3 + \frac{r\sqrt2}{2 - \frac{b^2}{a^2}}&= \frac{3a^2 - \frac{2r^2}{\left(2 - \frac{b^2}{a^2}\right)^2}}{a\sqrt3 - \frac{r\sqrt2}{2 - \frac{b^2}{a^2}}}\\&= \frac{3a^2 - \frac{2r^2}{\left(2 - \frac{b^2}{a^2}\right)^2}}{\frac{-rb}{2a - \frac{b^2}{a}}-c}\in \mathbb{Q} \end{align*}$$

However, if $a\sqrt3 + \frac{r\sqrt2}{2 - \frac{b^2}{a^2}} \in \mathbb{Q}$ and $a\sqrt3 - \frac{r\sqrt2}{2 - \frac{b^2}{a^2}} \in \mathbb{Q}$, then
$$\begin{align*} \left(a\sqrt3 + \frac{r\sqrt2}{2 - \frac{b^2}{a^2}}\right) + \left(a\sqrt3 - \frac{r\sqrt2}{2 - \frac{b^2}{a^2}}\right) = 2a\sqrt{3} \in \mathbb{Q} \end{align*}$$

This is a contradiction, since we assumed $a \neq 0$.

As you see the argument is very tedious (and perhaps there are even holes / mistakes, in which case please let me know). Is there a better (cleaner) way to prove the statement

Answer

Let $q = a\sqrt{6}+b\sqrt{3}+c\sqrt{2}$ with the premise that $a, b, c, q \in \mathbb{Q}$ ,then if $q = 0 \implies (-a\sqrt{6})^2 = (b\sqrt{3}+c\sqrt{2})^2\implies 6a^2 = 3b^2 + 2bc\sqrt{6} + 2c^2\implies \sqrt{6} \in \mathbb{Q}$, a contradiction unless $b = 0$ or $c = 0$ which in turn yields a much simpler case and a contradiction is easily obtained. If $q \neq 0$, then normalize the equation we can assume from now on $q = 1$ and repeat the steps above we have $\sqrt{6}$ is rational or $a = -bc$. The former can't happen and if the latter does, then we have: $1 = -bc\sqrt{6} + b\sqrt{3} + c\sqrt{2}\implies (b\sqrt{3} - 1)(1-c\sqrt{2}) = 0\implies \sqrt{3} = \dfrac{1}{b}$ or $\sqrt{2} = \dfrac{1}{c}$ which are rationals, and we have a contradiction again. Done.