Digitalize This

By completing the squares, the given equation can be rewritten as

$a^{2} + b^{2} + c^{2} = d(a + b + c) \Longrightarrow a^{2} - da + b^{2} - db + c^{2} - dc = 0 \Longrightarrow$

$\left(a - \dfrac{d}{2}\right)^{2} + \left(b - \dfrac{d}{2}\right)^{2} + \left(c - \dfrac{d}{2}\right)^{2} = 3 \times \dfrac{d^{2}}{4} \Longrightarrow (2a - d)^{2} + (2b - d)^{2} + (2c - d)^{2} = 3d^{2}$ .

So we need to look for 3 perfect squares, (keeping in mind that we could have squares of negative numbers), that sum to $3d^{2}$ for $d = 1, 2, ..., 9$ . Without loss of generality suppose $a \lt b \lt c$ . Now we can then have at most two of the perfect squares being the same, where either $2a - d = -(2b - d), 2b - d = -(2c - d)$ or $2a - d = -(2c - d)$ , (we cannot have all of the perfect squares the same as that would require 2 of $a,b,c$ to be equal to one another). We then proceed to look at the 9 cases separately:

• $d = 1 \Longrightarrow 3d^{2} = 3$ ; the only possibility here is $(2a - d)^{2}, (2b - d)^{2}, (2c - d)^{2} = (1,1,1)$ , which we can't have for reasons discussed above, so $d = 1$ is not possible;

• $d = 2 \Longrightarrow 3d^{2} = 12$ ; no combination (with up to 1 repeat) of $1,4,9$ add to $12$ , so $d \ne 2$ ;

• $d = 3 \Longrightarrow 3d^{2} = 27$ ; $(1,1,25)$ works, which gives us $(a,b,c) = (1,2,4)$ , and thus $d = 3$ is possible;

• $d = 4 \Longrightarrow 3d^{2} = 48$ ; no combination (with up to 1 repeat) of $1,4,9,16,25,36$ add to $48$ , so $d \ne 4$ ;

• $d = 5 \Longrightarrow 3d^{2} = 75$ ; the only possibility for the squares used is $(1, 25, 49)$ , which could be generated by either $(a,b,c) = (2,5,6)$ or $(3,5,6)$ , for which $b = d$ , and so $d \ne 5$ ;

• $d = 6 \Longrightarrow 3d^{2} = 108$ ; the squares $(4,4,100)$ follow from $(a,b,c) = (2,4,8)$ , so $d = 6$ is possible;

• $d = 7 \Longrightarrow 3d^{2} = 147$ ; the squares $(1,25,121)$ follow from $(a,b,c) = (4,6,9)$ , so $d = 7$ is possible:

• $d = 8 \Longrightarrow 3d^{2} = 192$ ; one of the squares used must be one of $81, 100, 121, 144$ or $169$ , but as none of $111, 92, 71, 48$ or $23$ is the sum of two squares we see that $d \ne 8$ ;

• $d = 9 \Longrightarrow 3d^{2} = 243$ ; the only possibility for the squares used is $(1,121,121)$ , but this would require that $b,c$ be $(-1,10)$ , which is not allowed, so $d \ne 9$ .

The only possible values for $d$ are thus $3,6,7$ , their sum being $\boxed{16}$ .

Note: While some brute force is used here, there are a limited number of squares to work with so the casework only took about 5 minutes, and thus I felt my approach was still worth sharing.

From the given equation, it is clear that $a+b+c| a^2+b^2+c^2$ . Hence, $a^2+b^2+c^2 \equiv 0 \pmod{a+b+c}$ .

And $(a+b+c)^2 \equiv (a^2+b^2+c^2)+2(ab+bc+ca) \equiv 0 \pmod{a+b+c}$ .

If $a^2+b^2+c^2 \equiv 0 \pmod{a+b+c}$ , then $2(ab+bc+ca) \equiv 0 \pmod{a+b+c}$ .

Then $2((a+b)(b+c) - b^2) \equiv 2((-c)(-a)-b^2) \equiv 2(ac - b^2) \equiv 0 \pmod{a+b+c}$ .

Therefore, if $b^2 = ac$ , the equivalence will be true, and from all the digit squares, here are the possibilities:

$2^2 = 1\cdot 4$ . Thus, $\dfrac{1^2 + 2^2 + 4^2}{1+2+4} = \dfrac{21}{7} = 3$ .

$3^2 = 1\cdot 9$ . Thus, $\dfrac{1^2 + 3^2 + 9^2}{1+3+9} = \dfrac{91}{13} = 7$ .

$4^2 = 2\cdot 8$ . Thus, $\dfrac{2^2 + 4^2 + 8^2}{2+4+8} = \dfrac{84}{14} = 6$ .

$6^2 = 4\cdot 9$ . Thus, $\dfrac{4^2 + 6^2 + 9^2}{4+6+9} = \dfrac{133}{19} = 7$ .

Now we have explored the most obvious solutions, but are there more?

$$ Since $b \equiv -a-c \pmod{a+b+c}$ , then $b^2 \equiv (a+c)^2 \pmod{a+b+c}$ .

Hence, $2(ac - b^2) \equiv 2(ac - (a+c)^2) \equiv 2(a^2 + ac + c^2) \equiv 0 \pmod{a+b+c}$ .

Then $2(c-a)(a^2 + ac + c^2) \equiv 2(c^3 - a^3) \equiv 0 \pmod{a+b+c}$ .

Similarly, with the terms alternated, we will have $2(c^3 - b^3) \equiv 2(b^3 - a^3) \equiv 0 \pmod{a+b+c}$ .

For generality, let us have $a<b<c$ . This way, we will attempt to find the middle digit for all possible square summation by checking the upper or the lower bound from that number.

Then the case of $b=2$ is already checked as shown in the above solution.

$$ For $b=3$ , we will have $a=2$ left, and $2(3^3 - 2^3) \equiv 2\cdot 19 \equiv 0 \pmod{2+3+c}$ . Then $c$ must be $14$ to make the equivalence true, but it's not the desired digit, according to the problem.

And we have the triple $a=1$ , $b=3$ , and $c=9$ as shown in the above solution.

$$ Then for $b=4$ , if $a=1$ , $2(4^3 - 1^3) \equiv 126 \equiv 2\cdot 3\cdot 3\cdot 7 \equiv 0 \pmod{1+4+c}$ . Then $c$ can be $4$ or $9$ , but if we want all distinct digits, only $9$ applies: $\dfrac{1^2 + 4^2 + 9^2}{1+4+9} = \dfrac{98}{14} = 7$ .

And we have the triple $a=2$ , $b=4$ , and $c=8$ as shown in the above solution.

For $b=4$ , if $a=3$ , $2(4^3 - 3^3) \equiv 2\cdot 37 \equiv 0 \pmod{1+4+c}$ . Then $c$ must be $32$ , which is not applicable.

$$ For $b=5$ , if $a=1$ , $2(5^3 - 1^3) \equiv 2\cdot 124 \equiv 8\cdot 31 \equiv 0 \pmod{1+5+c}$ . Only $2$ or $25$ works for $c$ here, but $2$ only works for $c=4$ as shown above, and $25$ is not applicable.

For $b=5$ , if $a=2$ , $2(5^3 - 2^3) \equiv 2\cdot 117 \equiv 2\cdot 3\cdot 3\cdot 13 \equiv 0 \pmod{2+5+c}$ . Only $2$ or $6$ works for $c$ , but if we want the distinct digit, only $6$ applies: $\dfrac{2^2 + 5^2 + 6^2}{2+5+6} = \dfrac{65}{13} = 5$ .

Unfortunately, it's not a distinct quotient digit, so not a new solution for us.

For $b=5$ , if $a=3$ , $2(5^3 - 3^3) \equiv 2\cdot 98 \equiv 2\cdot 2\cdot 7\cdot 7 \equiv 0 \pmod{3+5+c}$ . Only $6$ applies for $c$ : $\dfrac{3^2 + 5^2 + 6^2}{3+5+6} = \dfrac{70}{14} = 5$ . Again, it's not a distinct quotient digit, so not a new solution for us.

For $b=5$ , if $a=4$ , $2(5^3 - 4^3) \equiv 2\cdot 61 \equiv 0 \pmod{4+5+c}$ . No digit $c$ exists for this case.

$$ For $b=6$ , if $c=7$ , $2(7^3 - 6^3) \equiv 2\cdot 127 \equiv 0 \pmod{a+6+7}$ . No digit $a$ exists for this case.

For $b=6$ , if $c=8$ , $2(8^3 - 6^3) \equiv 2\cdot 296 \equiv 2\cdot 2\cdot 2\cdot 74 \equiv 0 \pmod{a+6+8}$ . No digit $a$ exists for this case.

And we have the triple $a=4$ , $b=6$ , and $c=9$ as shown in the above solution.

$$ For $b=7$ , if $c=8$ , $2(8^3 - 7^3) \equiv 2\cdot 169 \equiv 2\cdot 13\cdot 13 \equiv 0 \pmod{a+7+8}$ . No digit $a$ exists for this case.

For $b=7$ , if $c=9$ , $2(9^3 - 7^3) \equiv 2\cdot 386 \equiv 2\cdot 2\cdot 193 \equiv 0 \pmod{a+7+9}$ . No digit $a$ exists for this case.

$$ For $b=8$ , if $c=9$ , $2(9^3 - 8^3) \equiv 2\cdot 217 \equiv 2\cdot 7\cdot 31\equiv 0 \pmod{a+8+9}$ . No digit $a$ exists for this case.

$$

Thus, we can conclude that the only solution $(a,b,c)$ are: $(1,2,4)$ ; $(1,3,9)$ ; $(1,4,9)$ ; $(2,4,8)$ ; and $(4,6,9)$ .

And the only applicable quotient digits include: $3$ , $6$ , & $7$ .

Finally, the sum of all possible $d$ is $3+6+7 = \boxed{16}$ .

WLOG a<b<c. Used brute force to arrive at d=(3,5,6,7) but with 5, b=5. So we are left with 3+6+7=16.
Note that the values I have used for (a,b,c) are, keeping a, and b same let c=b+1 to 9. Next b=b+1, till b=9. THEN a=a+1, and repeat till (7,8,9).
(1,2,3)........(1,2,9)-------------7
(1,3,4).......(1,3,9)-------------6

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(1,8,9)--------------------------1

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(2,3,4,)......(1,3,9)--------------6

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(2,8,9)---------------------------1

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(7,8,9)-------------------------1

{(1+....7)+(1+....6)}+{(1+....5)+(1+....4)}+{(1+....3)+(1+2)}+(1)=\(7^2+5^2+3^2+1^2=84=7*4*(4-1) trials. ...... $\color{#D61F06}{Interesting~ Result.}$
once we get d as integer for some a, b and c, no need to take c to 9. Next star with b=b+1, and with new b, c=b+1.