Step 1: What kind of triangle is it.
$A(-4,9,k)$, $B(-1,k,k)$, $C(0,7,10)$ form an isosceles right triangle with $AB=BC$. So the equal sides $AB$ and $BC$ are the legs and $AC$ is the hypotenuse.
Step 2: Square the two equal legs.
\[ AB^2=(-1+4)^2+(k-9)^2+0^2=9+(k-9)^2=k^2-18k+90. \]
\[ BC^2=(0+1)^2+(7-k)^2+(10-k)^2=2k^2-34k+150. \]
Step 3: Set $AB=BC$.
$k^2-18k+90=2k^2-34k+150$ gives $k^2-16k+60=0$, i.e. $(k-6)(k-10)=0$, so $k=6$ or $k=10$.
Step 4: Use the integer-$AC$ condition.
For a right isosceles triangle, $AC^2=2AB^2$. With $k=6$: $AB^2=36-108+90=18$, so $AC^2=36$ and $AC=6$, an integer.
Step 5: Reject the other root.
With $k=10$: $AB^2=100-180+90=10$, so $AC^2=20$ and $AC=2\sqrt5$, not an integer. So $k=6$.
Step 6: Add up the perimeter.
With $k=6$, $AB=BC=\sqrt{18}=3\sqrt2$ and $AC=6$, so
\[ P=3\sqrt2+3\sqrt2+6=6\sqrt2+6=6(1+\sqrt2). \]
\[ \boxed{6(1+\sqrt2)} \]