Pandigital Sum whose Components are Multiples

Theorem
The following sums use each of the digits $1$ to $9$ exactly once each, while one summand equals $2$ times the other summand:

It is interesting to note how the $4$ sums can be divided into two pairs in which each element of the sum is an anagram of its corresponding element in the other.

Proof
We establish some parameters for the number $x$ such that the equation $x + 2 x = 3 x$ is pandigital.

Firstly, among the three numbers, there cannot be repeating digits or $0$.

If $5 \divides x$:
 * $10 \divides 2 x$ and $2 x$ ends in $0$.

By Divisibility by 5, $x$ cannot end in $5$.

Secondly, $x$ must be divisible by $3$.

To see this, consider the concatenation of the three numbers, $\sqbrk {x || 2 x || 3 x}$.

Since the digits of this number is a permutation of $\set {1, \dots, 9}$, they sum to $45$.

Since $9 \divides 45$, by Divisibility by 9:
 * $9 \divides \sqbrk {x || 2 x || 3 x}$

We have:

Hence $9 \divides 6 x$.

This gives $3 \divides 2 x$.

By Euclid's Lemma:
 * $3 \divides x$

Finally, as $3 x < 1000$, we must have $100 \le x \le 333$.

Putting repeating digits and zeroes into consideration, we list and check all possible candidates from $123$ to $327$:

Hence the result.