If $(x^2-4x+7)^3+p(x^2-4x+7)^2+q(x^2-4x+7)+r=0$ has no real roots, Then $p+2q+r$ is 
If $f(x)=x^3+px^2+qx+r$ has three distinct non negative integer roots and
$(x^2-4x+7)^3+p(x^2-4x+7)^2+q(x^2-4x+7)+r=0$ has no real roots,
Then $p+2q+r$ is

$\bf{Attempt:}$ Assuming  $f(x) = x^3+px^2+qx+r = (x-a)(x-b)(x-c)$,
where $a,b,c \geq 0$ and $a,b,c \in \mathbb{Z}$
Then  $f(x^2-4x+7) = (x^2-4x+7)^3+p(x^2-4x+7)^2+q(x^2-4x+7)+r $
$=(x^2-4x+7-a)(x^2-4x+7-b)(x^2-4x+7-c)$
Could some help me how to solve it , thanks

$$
f(x^2-4x+7)=((x-2)^2+(3-a))((x-2)^2+(3-b))((x-2)^2+(3-c))
$$
has no real roots so $3-a$, $3-b$, and $3-c$ are at least $1$. Hence
$$
\{a,b,c\}=\{0,1,2\}.
$$
