Proof that $\sqrt6 - \sqrt2 - \sqrt3$ is irrational.
Suppose $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is rational.
Then, $(\sqrt{3}-1)(\sqrt{2}-1)=\sqrt{6}-\sqrt{2}-\sqrt{3}+1$ is a rational number, say $r\in\mathbb{Q}$.
That is, $\sqrt{3}-1=\frac{r}{\sqrt{2}-1}=r(\sqrt{2}+1)$.
Thus, $\sqrt{3}-r\sqrt{2}=r+1\in\mathbb{Q}$.
Clearly, $r\neq -1$, whence $\sqrt{3}-r\sqrt{2}\neq 0$.
Now, $\sqrt{3}+r\sqrt{2}=\frac{3-2r^2}{\sqrt{3}-r\sqrt{2}}=\frac{3-2r^2}{r+1}\in\mathbb{Q}$.
What happens if both $\sqrt{3}-r\sqrt{2}$ and $\sqrt{3}+r\sqrt{2}$ are rational numbers?
This line of reasoning shows that $a\sqrt{pq}+b\sqrt{p}+c\sqrt{q}$ is irrational if $a,b,c\in\mathbb{Q}$ with $a\neq0$ and $p,q\in\mathbb{N}\setminus\{1\}$ are such that $p$ and $q$ are distinct and square-free.
I think the direct method is fairly simple, really. Not sure it adds anything to the other answers at this point, but just in case.
NOTE: a commenter helpfully pointed out an algebraic error in the original version of this solution. This error had no significant impact on the solution, and it has now been corrected.
Suppose $\sqrt 6 -\sqrt 2 - \sqrt 3$ were rational. Square to see that $$6 + 2 + 3- 4\sqrt 3 - 6\sqrt 2 + 2\sqrt 6 \;\; \in \mathbb Q$$ Which implies that: $$ \sqrt 6 - 2\sqrt 3 - 3\sqrt 2 \;\; \in \mathbb Q$$ Subtracting this from the original expression we see that $$2 \sqrt 2 + \sqrt3 \;\; \in \mathbb Q$$ Square again and simplify to deduce that $$\sqrt 6 \in \mathbb Q$$ which is false, giving us the contradiction we sought.
The brute-force method, for when no clever argument such as in the other answer applies, would be something like:
Let $X=\sqrt{6}-\sqrt2-\sqrt3$.
Calculate $1$, $X$, $X^2$, $X^3$, $X^4$ as rational linear combinations of $1$, $\sqrt2$, $\sqrt3$ and $\sqrt 6$.
Because the expressions for these 5 powers of $X$ lie in a 4-dimensional vector space over $\mathbb Q$, they must have a nontrivial linear relation, that is, a degree-4 polynomial with rational coefficients that has $X$ as a root. Find such a polynomial using linear algebra.
Appply the rational root theorem to see if the polynomial has any rational roots. If not, then $X$ cannot be rational.