Question 1205480
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This is a system of 2 equations with 3 unknowns.
The fact we have more unknowns than equations leads to "infinitely many solutions" for this system.
It turns out that each solution is of the form (a,b,c) = (-9-c,6+c,c) which I explain in a later section below. 


Let's say c = 0
b-c = 6
b-0 = 6
b = 6
Then,
a+b = -3 
a+6 = -3 
a = -3-6 
a = -9
Or you could say
(a,b,c) = (-9-c,6+c,c)
(a,b,c) = (-9-0,6+0,0)
(a,b,c) = (-9,6,0)


Therefore,
{{{2a^2 - 3b^2 + c^2 = 2(-9)^2 - 3(6)^2 + (0)^2 = 54}}}


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Another example


Let c = 1
b-c = 6
b-1 = 6
b = 6+1
b = 7
Then,
a+b = -3 
a+7 = -3 
a = -3-7 
a = -10
Or
(a,b,c) = (-9-c,6+c,c)
(a,b,c) = (-9-1,6+1,1)
(a,b,c) = (-10,7,1)


Therefore,
{{{2a^2 - 3b^2 + c^2 = 2(-10)^2 - 3(7)^2 + (1)^2 = 54}}}


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One more example


Let's say c = 2
b-c = 6
b-2 = 6
b = 6+2
b = 8
Then,
a+b = -3 
a+8 = -3 
a = -3-8 
a = -11
Or
(a,b,c) = (-9-c,6+c,c)
(a,b,c) = (-9-2,6+2,2)
(a,b,c) = (-11,8,2)


Therefore,
{{{2a^2 - 3b^2 + c^2 = 2(-11)^2 - 3(8)^2 + (2)^2 = 54}}}


It appears we keep landing on 54. 
Is this a coincidence? Or is this always going to happen? 
The next section will shed light on that.


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A more generalized approach.


b-c = 6
b = 6+c
a+b = -3
a+(6+c) = -3
a = -3-6-c
a = -9-c


We have
a = -9-c
b = 6+c
c = c
in which we can say
(a,b,c) = (-9-c,6+c,c)
This confirms that the system a+b = -3 and b-c = 6 has infinitely many solutions.



So,
{{{2a^2 - 3b^2 + c^2 = 2(-9-c)^2 - 3(6+c)^2 + c^2}}}


{{{2a^2 - 3b^2 + c^2 = 2(c^2+18c+81) - 3(c^2+12c+36) + c^2}}}


{{{2a^2 - 3b^2 + c^2 = 2c^2+36c+162 -3c^2-36c-108 + c^2}}}


{{{2a^2 - 3b^2 + c^2 = (2c^2 -3c^2 + c^2)+(36c-36c)+(162-108)}}}


{{{2a^2 - 3b^2 + c^2 = 0c^2+0c+54}}}


{{{2a^2 - 3b^2 + c^2 = 54}}}
This proves that if a+b = -3 and b-c = 6, then {{{2a^2 - 3b^2 + c^2}}} will always land on 54.


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Answer: <font color=red size=4>54</font>
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