Real values!

Let $x_i-3=r_i$ for $i=1,2,3$

$r_1,r_2,r_3$ are the roots of ${ (x+3) }^{ 3 }-{ 6(x+3) }^{ 2 }+a(x+3)-a=0$

$\Rightarrow x^3+3x^2+(a-9)x+2a-27=0$

Using Vieta,

$r_1+r_2+r_3=-3$

$r_1r_2+r_2r_3+r_3r_1= a-9$

$r_1r_2r_3=27-2a$

We know,

$r_1^3+r_2^3+r_3^3-3r_1r_2r_3=(r_1+r_2+r_3)(r_1^2+r_2^2+r_3^2-r_1r_2-r_2r_3-r_3r_1)$

$\Rightarrow r_1^3+r_2^3+r_3^3-3r_1r_2r_3=(r_1+r_2+r_3)\{(r_1+r_2+r_3)^2-3(r_1r_2+r_2r_3+r_3r_1)\}$

$\Rightarrow r_1^3+r_2^3+r_3^3=-3\{9-3(a-9)\}+3(27-2a)=0 \ [Given]$

Solving this , $-a=\fbox{-9}$

Shouldn't be a level 5

$x^{3}-6x^{2}+ax-a=0$ $\Rightarrow$ given

$(x^3-9x^2+27x-27)+3x^2+(a-27)x-(a-27)=0$

$(x-3)^3+3x^2+(a-27)x-(a-27)=0$

$\sum_{i=1}^3(x_{i}-3)^3+3x_{i}^2+(a-27)x_{i}-(a-27)=0$ where, $x_{1},x_{2},x_{3}$ are roots

$\sum_{i=1}^3(x_{i}-3)^3=0$ $\Rightarrow$ given

$\Rightarrow$ $(3(\sum_{i=1}^3x_{i}^2)+((a-27)\sum_{i=1}^3x_{i})-((a-27)\sum_{i=1}^31)=0$

$\sum_{i=1}^3x_{i}=6$ $\Rightarrow$ sum of roots

$\sum_{i=1}^3x_{i}^2=(\sum x_{i})^2-2(x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{1})=36-2a$

substituting these values we get

$3(36-2a)+(a-27)6-(a-27)3=0$

$\Rightarrow$ $a=9$ or $-a=-9$