Does $a^3 + 2b^3 + 4c^3 = 6abc$ have solutions in $\mathbb{Q}$

First, note that $(a,b,c)$ is a solution if and only if $(ka,kb,kc)$ is; hence we may assume $a,b,c$ are integers, with no common factor (divide by that common factor if necessary).

Because $6abc, 2b^3+4c^3$ are even, so is $a^3$ and hence $a$. Write $a=2a'$ and we have $$8(a')^3+2b^3+4c^3=12a'bc$$ and hence $$4(a')^3+b^3+2c^3=6a'bc$$

By similar logic, $b$ is even, so write $b=2b'$ and we have $$4(a')^3+8(b')^3+2c^3=12a'b'c$$ But now $$2(a')^2+4(b')^3+c^3=6a'b'c$$ and hence $c$ is even. Hence $a,b,c$ are all even; this contradicts $a,b,c$ having no common factor.

Followup: The same proof works if the coefficients $\{1,2,4,6\}$ are replaced by $\{\alpha_1, \alpha_2, \alpha_3,\alpha_4\}$ so long as there is some prime $p$ with $\nu_p(\alpha_1)=0, \nu_p(\alpha_2)=1, \nu_p(\alpha_3)=2, \nu_p(\alpha_4)\ge 1$. (Here $\nu_p(\cdot)$ denotes the p-adic valuation). For example, apart from $(0,0,0)$, there are no rational solutions to $$7a^3+15b^3+18c^3=45abc$$ where here $p=3$.

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Double followup: The same proof works with $n$ variables $$\alpha_0a_0^n+\alpha_1a_1^n+\cdots+\alpha_{n-1}a_{n-1}^n=\alpha_n(a_0a_1\cdots a_{n-1})$$ provided that $\nu_p(\alpha_i)=i$ (for $0\le i\le n-1$) and $\nu_p(\alpha_n)\ge 1$.

As it happens, the cubic form $C(a,b,c)=a^3+2b^3+4c^3-6abc$ is the norm form for the extension $K=\mathbb Q(\root 3 \of 2)$ over $\mathbb Q$. That is, if you look at a general element of $K$, say $a + b\root3\of2+c\root3\of2^2$, and take its field-theoretic norm, what you get is exactly $C(a,b,c)$. Now, the norm doesn’t vanish on an algebraic extension, except at zero. So, just because $C$ happens to be a norm form, you can say immediately that the trivial zero is the only one.

How did I spot this? By having done lots of examples, first by hand over many years, then, more recently, with symbolic algebra programs.

HINT Make use of $$x^3 + y^3 + z^3 -3xyz = \left(x+y+z \right) \left(x^2+y^2+z^2-xy-yz-zx \right)$$where $x=a$, $y=b\sqrt[3]{2}$ and $z=c \sqrt[3]{4}$.