Correct Answer: $6\sqrt{6}$

Solution : Given: $x=1+\sqrt{2}+\sqrt{3}$
We know that the algebraic identities are $(x-y)^2=x^2+y^2-2xy$ and $(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx$.
$x-1=\sqrt{2}+\sqrt{3}$
On squaring both sides of the above equation, we get,
$(x-1)^2=(\sqrt{2}+\sqrt{3})^2$
⇒ $(x^2+1-2x)=(3+2+2\sqrt{6})$
⇒ $(x^2-2x-4)=2\sqrt{6}$ (equation 1)
On squaring both sides of the above equation, we get,
$(x^2-2x-4)^2=(2\sqrt{6})^2$
⇒ $x^4+4x^2+16-4x^3+16x-8x^2=24$
⇒ $x^4-4x^3-4x^2+16x-8=0$
Multiply by 2 on both sides of the above equation, we get,
$2(x^4-4x^3-4x^2+16x-8)=0$
⇒ $2x^4-8x^3-8x^2+32x-16=0$
⇒ $2x^4-8x^3-5x^2+26x-28-3x^2+6x+12=0$
⇒ $2x^4-8x^3-5x^2+26x-28=3(x^2-2x-4)$
From equation 1, substitute the value in the above equation and we get,
$(2x^4-8x^3-5x^2+26x-28)=3\times2\sqrt{6}$
⇒ $(2x^4-8x^3-5x^2+26x-28)=6\sqrt{6}$
Hence, the correct answer is $6\sqrt{6}$.