Solutions to $100a+10b+c=11a^2+11b^2+11c^2$ and $100a+10b+c=11a^2+11b^2+11c^2+ k$
We want to find all sets of integers $a,b,c,k$ such that $$100a+10b+c=11a^2+11b^2+11c^2+ k\tag1$$ where $1\le a\le 9,0\le b\le 9,0\le c\le 9$ and $0\le k\le 10$.
We have $$(1)\iff k+b(11b-10)+c(11c-1)=a(100-11a)$$
Since $a(100-11a)\le 5(100-11\times 5)=225$, we have $$k+b(11b-10)+c(11c-1)\le 225\tag2$$
Since we have that $b(11b-10)\ge 6(11\times 6-10)=336\gt 225$ for $b\ge 6$ and that $c(11c-1)\ge 5(11\times 5-1)=270\gt 225$ for $c\ge 5$, from $(2)$, we have to have $b\le 5$ and $c\le 4$.
Also, from $(1)$ with $0\le k\le 10$, solving $$0\le 100a+10b+c-11a^2-11b^2-11c^2\le 10$$for $a$ gives$$a\in\left[\frac{50-\sqrt m}{11},\frac{50-\sqrt{m-110}}{11}\right]\cup\left[\frac{50+\sqrt{m-110}}{11},\frac{50+\sqrt m}{11}\right]\tag3$$where $m=2500-11(11b^2+11c^2-10b-c)$.
Now, we see that $$\frac{50-\sqrt m}{11}\le 1\le \frac{50-\sqrt{m-110}}{11}\iff 1521\le m\le 1631\quad\text{for}\quad a=1$$$$\frac{50-\sqrt m}{11}\le 2\le \frac{50-\sqrt{m-110}}{11}\iff 784\le m\le 894\quad \text{for}\quad a=2$$$$\frac{50-\sqrt m}{11}\le 3\le \frac{50-\sqrt{m-110}}{11}\iff 289\le m\le 399\quad\text{for}\quad a=3$$$$\frac{50-\sqrt m}{11}\le 4\le\frac{50-\sqrt{m-110}}{11}\iff 110\le m\le 146\quad\text{for}\quad a=4$$$$\frac{50+\sqrt{m-110}}{11}\le 5\le\frac{50+\sqrt m}{11}\iff 100\le m\le 125\quad \text{for}\quad a=5$$$$\frac{50+\sqrt{m-110}}{11}\le 6\le\frac{50+\sqrt m}{11}\iff 256\le m\le 356\quad\text{for}\quad a=6$$$$\frac{50+\sqrt{m-110}}{11}\le 7\le\frac{50+\sqrt m}{11}\iff 729\le m\le 829\quad\text{for}\quad a=7$$$$\frac{50+\sqrt{m-110}}{11}\le 8\le \frac{50+\sqrt m}{11}\iff 1444\le m\le 1544\quad\text{for}\quad a=8$$$$\frac{50+\sqrt{m-110}}{11}\le 9\le\frac{50+\sqrt m}{11}\iff 2401\le m\le 2501\quad\text{for}\quad a=9$$
Case 1 : When $b=5$, since $b(11b-10)=225$ with $(2)$, we have to have $k=c(11c-1)=0,a=5$ giving $a=5,b=5,c=0,k=0$.
Case 2 : When $b=4$, since $b(11b-10)=136$, $$(2)\implies k+c(11c-1)\le 225-136=89$$from which we have to have $c\le 2$ since $c(11c-1)\ge 3\times (11\times 3-1)=96\gt 89$ for $c\ge 3$.
Case 2-1 : For $c=0$, we have $m=1004$. So, there is no integer $a$ satisfying $(3)$.
Case 2-2 : For $c=1$, we have $m=894$. So, we have $a=2$, and $k=10$ from $(1)$.
Case 2-3 : For $c=2$, we have $m=542$. So, there is no integer $a$ satisfying $(3)$.
Case 3 : When $b=3$, since $b(11b-10)=69$, $$(2)\implies k+c(11c-1)\le 225-69=156$$from which we have to have $c\le 3$ since $c(11c-1)\ge 4\times (11\times 4-1)=172\gt 156$ for $c\ge 4$.
Case 3-1 : For $c=0$, we have $m=1741$. So, there is no integer $a$ satisfying $(3)$.
Case 3-2 : For $c=1$, we have $m=1631$. So, we have $a=1$, and $k=10$ from $(1)$.
Case 3-3 : For $c=2$, we have $m=1279$. So, there is no integer $a$ satisfying $(3)$.
Case 3-4 : For $c=3$, we have $m=685$. So, there is no integer $a$ satisfying $(3)$.
Case 4 : When $b=2$, we have $c\le 4$.
Case 4-1 : For $c=0$, we have $m=2236$. So, there is no integer $a$ satisfying $(3)$.
Case 4-2 : For $c=1$, we have $m=2126$. So, there is no integer $a$ satisfying $(3)$.
Case 4-3 : For $c=2$, we have $m=1774$. So, there is no integer $a$ satisfying $(3)$.
Case 4-4 : For $c=3$, we have $m=1180$. So, there is no integer $a$ satisfying $(3)$.
Case 4-5 : For $c=4$, we have $m=344$. So, we have $a=3,6$, and $(a,b,c,k)=(3,2,4,5),(6,2,4,8)$ from $(1)$.
Case 5 : When $b=1$, we have $c\le 4$.
Case 5-1 : For $c=0$, we have $m=2489$. So, we have $a=9$ and $k=8$ from $(1)$.
Case 5-2 : For $c=1$, we have $m=2379$. So, there is no integer $a$ satisfying $(3)$.
Case 5-3 : For $c=2$, we have $m=2027$. So, there is no integer $a$ satisfying $(3)$.
Case 5-4 : For $c=3$, we have $m=1433$. So, there is no integer $a$ satisfying $(3)$.
Case 5-5 : For $c=4$, we have $m=597$. So, there is no integer $a$ satisfying $(3)$.
Case 6 : When $b=0$, we have $c\le 4$.
Case 6-1 : For $c=0$, we have $m=2500$. So, we have $a=9$, and $k=9$ from $(1)$.
Case 6-2 : For $c=1$, we have $m=2390$. So, there is no integer $a$ satisfying $(3)$.
Case 6-3 : For $c=2$, we have $m=2038$. So, there is no integer $a$ satisfying $(3)$.
Case 6-4 : For $c=3$, we have $m=1444$. So, we have $a=8$, and $k=0$ from $(1)$.
Case 6-5 : For $c=4$, we have $m=608$. So, there is no integer $a$ satisfying $(3)$.
Therefore, the answer is $$\begin{align}\color{red}{(a,b,c,k)=}\ &\color{red}{(5,5,0,0),(2,4,1,10),(1,3,1,10),(3,2,4,5)}\\&\color{red}{(6,2,4,8),(9,1,0,8),(9,0,0,9),(8,0,3,0)}\end{align}$$
Make a table of all numbers, multiples of 11 from 110 to 990 as below:
There are two cases which satisfy the solution: 550 & 803. I did this simulation in excel.
n a b c a^2 b^2 c^2 sum n/11 f
550 5 5 0 25 25 0 50 50 1
803 8 0 3 64 0 9 73 73 1
EDIT: If I don't assume that all numbers would be multiples of 11, there are 8 solutions:
n a b c sum n/11 qoutient
131 1 3 1 11 11
241 2 4 1 21 21
324 3 2 4 29 29
550 5 5 0 50 50
624 6 2 4 56 56
803 8 0 3 73 73
900 9 0 0 81 81
910 9 1 0 82 82
This might not be the best way but here is how I would do it.
Assume the remainder to be $\lambda$. Remember that $\lambda$ can take any value from 0 to 10.
Note that $100a + 10b +c = 11a^2 + 11b^2 + 11c^2 + \lambda$ can be rewritten as
$$(11a - 50)^2 + (11b - 5)^2 + (11c - \frac{1}{2})^2 = 2525 + \frac{1}{4} - \lambda$$
For the remainder of the answer, I'll assume $\lambda = 0.$ The other cases for $\lambda$ can be handled similarly. Since each of the terms on the left hand side are positive, we get that $c \leq 4$
We can take 5 cases based upon each of the five possible values for $c$ and solve for $a$ and $b$.
I won't go through all of the cases but in each one of them we should be able to form upper bounds on the values of $a$ and $b$. Furthermore, the search space can be pruned even further by making use of the parity of $a$ and $b$.
This method is not very different from your "table generating" method. The only difference is that the search space is pruned quite a bit. Also, if the number is itself divisible by 11, we can use the property mentioned in the comments to prune the search space even more.
The process can be repeated for other values of $\lambda$. The upper bounds should not differ a lot but the individual solutions will.