Let the three digit number be $N=100a+10b+c$, where $a$, $b$, $c$ are digits, $1 \leq a\leq9$, and $0 \leq b, c\leq 9$.

We are given two conditions:

1. $N$ is divisible by $11$.
2. $\frac{N}{11}=a^2+b^2+c^2 \implies N=11\left(a^2+b^2+c^2\right)$.

By the divisibility rule for $11$, the alternating sum of digits $a-b+c$ must be a multiple of $11$. Since $1 \leq a\leq9$ and $0 \leq b, c\leq 9$, the value of $a-b+c$ can only be $0$ or $11$.

Case 1:

Let $b=a+c.$

Then, we get:

Since the right side is even, the left side must also be even, meaning that $c$ must be an even digit ($0, 2, 4, 6, 8$).

If $c=0$, then

Then,

$b=5+0$

This gives the number $N=505$.

Check: $\frac{550}{11}=50$ and $5^2+5^2+0^2=50$. So, it is valid.

Testing $c=2, 4, 6, 8$ yields quadratic equations with no integer solutions for $a$.

Case 2:

Let $b=a+c-11$.

Since $b\geq0$, we must have $a+c\geq11$. Substituting $b$ into the expression of $N$:

Equating this to our second condition:

Let’s test the possible sums for $a+c$ from $11$ to $18$:

If $a+c=11$, then $b=0.$

Substitute $c=11-a$.

Solving this quadratic equation using the quadratic formula:

When $a=8$, $c=11-8=3$. This gives the number $N=803$.

Check: $\frac{803}{11}=73$ and $8^2+0^2+3^2=73$. So, it is valid.

Testing other values for $a+c$ from $12$ to $18$ yields no other integer solutions.

The only three-digit numbers satisfying the property are $550$ and $803$.