Given, the points $A\left(1,3,5\right), B\left(2,4,6\right),C\left(4,5,k\right)$ form a right angled triangle.

Now, $A{B}^2=1^2+1^2+1^2=3$

$B{C}^2=2^2+1^2+{\left(k-6\right)}^2$$=k^2-12k+41$ and

$C{A}^2=3^2+2^2+{\left(k-5\right)}^2$$=k^2-10k+38$

If right angled at $A,$ then

$A{B}^2+A{C}^2=B{C}^2$

$\Rightarrow 3+k^2-10k+38=k^2-12k+41$

$\Rightarrow k=0$

If right angled at $B,$ then

$A{B}^2+B{C}^2=A{C}^2$

$\Rightarrow 3+k^2-12k+41=k^2-10k+38$

$\Rightarrow k=3$

If right triangled at $C,$ then

$A{C}^2+B{C}^2=A{B}^2$

$\Rightarrow k^2-12k+41+k^2-10k+38=3$

$\Rightarrow k^2-11k+38=0$

Not possible