Here's some sound and efficient algebra from Conor from Queen Elizabeth's Hospital :

We have three numbers, $x, y$ and $z$.

$x + y = 11 \quad (1)$
$y + z = 17 \quad (2) $
$z + x = 22 \quad (3) $

Adding $(1)$ and $(2)$ gives :

$x + 2y + z = 28 $

Putting that another way :

$(x + z) + 2y = 28 \quad (4) $

Substituting $(3)$ into $(4)$ gives :

$22 + 2y = 28 $

So $y = 3$.

Substituting $y = 3$ into $(1)$ gives $x = 8$ and substituting it into $(2)$ gives $z = 14$.


But Eden has had an excellent approach too :

$11 + 17 + 22$ uses every number twice and makes a total of $50$

So the total of the three numbers that we need to find is $25$.

If two of them make $11$ together the one not used must have been $14$.

Two together made $17$, the one not used this time must have been $8$.

And finally, a pair have a sum of $22$, so the other number is $3$.

The three numbers are $3, 8,$ and $14$

And Mark used some deductive reasoning combined with an exhaustive approach:

The smallest number must be between $1$ and $5$ to make a smallest sum of $11$.
If it is $1$ then to make $11$ the second number is $10$.
If the third number is $x$
Then $x+1= 17$
and $x+10 = 22$
This does not work.

I tried the smallest number as $2$, then the second number is $9$
Then $x+2= 17$ and $x+9 = 22$
This also does not work but the two values of $x$ are closer than last time so I think $3$ will work.

Trying the first number as $3$ and the second as $8$
Then $x+3= 17$ and $x+8 = 22$
This works.

The three numbers are $3, 8$ and $14$.