Beautiful Quadrilateral Identity

$\begin{array}{lll} a^2+b^2+c^2&=&{(a+b+c)}^2-2(ab+bc+ac) \\ &=&{((c+6)+c)}^2-2((bc+ac-1)+bc+ac) \\ &=&{(2c+6)}^2-2(2bc+2ac-1) \\ &=&{(4c^2+24c+36)}-2(2c(b+a)-1) \\ &=&{(4c^2+24c+36)}-2(2c(c+6)-1) \\ &=&{(4c^2+24c+36)}-(4c^2+24c-2) \\ &=& \boxed{38} \end{array}$

$({ a }+{ b }-{ c })^{ 2 } = 36\\ a^{ 2 }+b^{ 2 }+c^{ 2 } = 36-2ab+2bc+2ac\\ = 2(18-ab+bc+ac)\\ = 2(18-ab+(ab-ac+1)+ac)\\ = 2(19) = 38$

Use the identity

(a+b-c)^2 = a^2 + b^2 + c^2 + 2( ab -bc-ca )

Rearranging the equations,

a + b - c = 6 $\boxed{1}$

ab - ac -bc = -1 $\boxed{2}$

Squaring equation $\boxed{1}$ :

$a^2+b^2+c^2 + 2(ab-ac-bc) = 36$ $\boxed{1'}$ (using the $(a+b-c)^2$ identity)

Substituting equation $\boxed{2}$ into equation $\boxed{1'}$ ,

$a^2+b^2+c^2 + 2(-1) = 36$

$a^2+b^2+c^2 - 2 = 36$

$a^2+b^2+c^2 = \boxed{38}$