The 3 cute roots

$Dividing \ the \ equation\ by\ \ i,\ \ z^3-6iz^2-11z+6i=0.\\ By\ Vieta's\ Formula,\ a+b+c=i6,\ \ ab+bc+ca=-11.\\ \therefore\ (a+b+c)^2=-36=a^2+b^2+c^2+2(ab+bc+ca)=a^2+b^2+c^2+2(- 11).\\ \implies\ -36+22={\Large\ \color{#D61F06}{-14}}=a^2+b^2+c^2\\ \ \ \ \\$ .

OR

$Dividing \ the \ equation\ by\ \ i,\ \ f(z)= z^3-6iz^2-11z+6i=0.\\ By\ Vieta's\ Formula,\ a+b+c=i6.\ \ abc = - 6i.\\ f(i)=0 \ \therefore a=i.\ \ but\ \ abc= - 6i.\\ \therefore\ bc=-6.\ \ \implies\ try \ b=2i,\ \ c=3i.\\ f(2i)=0,\ \ and\ \ f(3i)=0.\\ \therefore\ \ a^2+b^2+c^2=-1 - 4 - 9=-14.\\ If \ try\ had\ failed\ next\ alternative\ was\ b=-2,\ c=3\ \ \ or\\ b=2,\ c=-3.$

$i(z-a)(z-b)(z-c)=P(x)=i{ z }^{ 3 }+6{ z }^{ 2 }-11iz-6\\ -iabc=P(0)=-6....\quad abc=-6i\\ Let\quad d\quad be\quad the\quad common\quad difference\quad of\quad the\quad AP,\\ a+b+c=6i\quad \quad \quad \quad c=a+2d\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad b=a+d\\ 3a+3d=6i\quad so,\quad a=2i-d\Longrightarrow \quad b=2i\\ so,\quad c=4i-a\quad ....\quad a(2i)(4i-a)=-6i\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -2i{ a }^{ 2 }-8a+6i=0\\ \Delta =64-48=16\\ a=\frac { 8+4 }{ -4i } =3i\quad or\quad a=\frac { 8-4 }{ -4i } =i\\ Take\quad a=i\quad so\quad c=3i\quad and\quad b=2i\\ (it's\quad an\quad AP\quad of\quad common\quad differnce=d=i)\\ \\ \\ { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }=-1-4-9=-14.\\ \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\$