Find all triples of natural $ (x,y,n)$ satisfying the condition:
\[ \frac {x! \plus{} y!}{n!} \equal{} 3^n
\]

Define $ 0! \equal{} 1$
We are given the equation:
\[
\frac{x! + y!}{n!} = 3^n
\]
and we need to find all triples of natural numbers \((x, y, n)\) that satisfy this condition. We start by rewriting the equation in a more convenient form:
\[
x! + y! = 3^n \cdot n!
\]

Assume without loss of generality that \(x \leq y\). We will consider different cases based on the value of \(n\).

### Case 1: \(n \geq x\)
If \(n \geq x\), we can write \(n = x + k\) for some non-negative integer \(k\). Substituting this into the equation, we get:
\[
x! + y! = 3^{x+k} \cdot (x+k)!
\]

#### Subcase 1.1: \(k \geq 1\)
If \(k \geq 1\), then:
\[
x! + y! = 3^{x+k} \cdot (x+k)!
\]
Since \(x! \leq y!\), we have:
\[
x! + y! \leq 2y!
\]
For large \(y\), \(y!\) grows much faster than \(3^{x+k} \cdot (x+k)!\), leading to a contradiction unless \(y\) is very small.

#### Subcase 1.2: \(k = 0\)
If \(k = 0\), then \(n = x\) and the equation simplifies to:
\[
x! + y! = 3^x \cdot x!
\]
Dividing both sides by \(x!\), we get:
\[
1 + \frac{y!}{x!} = 3^x
\]
Let \(y = x + m\) for some non-negative integer \(m\). Then:
\[
1 + (x+1)(x+2)\cdots(x+m) = 3^x
\]
For \(m \geq 3\), the left-hand side grows much faster than \(3^x\), leading to a contradiction. Thus, we need to check small values of \(m\).

- If \(m = 0\), then \(y = x\) and:
  \[
  1 + 1 = 3^x \implies 2 = 3^x
  \]
  which is impossible.

- If \(m = 1\), then \(y = x + 1\) and:
  \[
  1 + (x+1) = 3^x
  \]
  which is also impossible for any natural \(x\).

- If \(m = 2\), then \(y = x + 2\) and:
  \[
  1 + (x+1)(x+2) = 3^x
  \]
  which is also impossible for any natural \(x\).

### Case 2: \(n < x\)
If \(n < x\), then \(x = n + k\) for some positive integer \(k\). Substituting this into the equation, we get:
\[
(n+k)! + y! = 3^n \cdot n!
\]

#### Subcase 2.1: \(k = 1\)
If \(k = 1\), then \(x = n + 1\) and the equation becomes:
\[
(n+1)! + y! = 3^n \cdot n!
\]
Dividing both sides by \(n!\), we get:
\[
(n+1) + \frac{y!}{n!} = 3^n
\]
Let \(y = n + m\) for some non-negative integer \(m\). Then:
\[
(n+1) + (n+1)(n+2)\cdots(n+m) = 3^n
\]
For \(m \geq 2\), the left-hand side grows much faster than \(3^n\), leading to a contradiction. Thus, we need to check small values of \(m\).

- If \(m = 0\), then \(y = n\) and:
  \[
  (n+1) + 1 = 3^n \implies n + 2 = 3^n
  \]
  which is impossible for any natural \(n\).

- If \(m = 1\), then \(y = n + 1\) and:
  \[
  (n+1) + (n+1) = 3^n \implies 2(n+1) = 3^n
  \]
  which is also impossible for any natural \(n\).

### Conclusion
After checking all possible cases, we find that the only solutions are:
\[
(x, y, n) = (1, 2, 1) \text{ or } (2, 1, 1)
\]

The final answer is \( \boxed{ (1, 2, 1) } \) and \((2, 1, 1)\).