WLOG $(x,y,z,w)=1$ If $z=2a$ then with mod8 we have: $w^2+3x^2=0(mod8)$ which gives $w=4b$ and $x=4c$ Now the equation become:$16b^2+16*11c^2-8y^2-24ya-40a^2=0$ So $2b^2+2*11c^2-y^2-3ya-5a^2=0$ Which gives $y^2+3ya+5a^2=even$ which mean both $a,y=even$ Contradiction because $(x,y,z,w)=1$ . Therefore $z=odd$. If $y=even$ then mod 8 gives: $w^2+11x^2-10z^2=w^2+3x^2-2=0(mod8)$ so $w^2+3x^2=2(mod8)$ contradiction. So $y,z=odd$now mod8 gives: $w^2+3x^2-4-2=0(mod8)$ obviously no solution. So the problem has now solution. Edit: we have lost the solution$(x,y,z,w):(0,0,0,0)$ because in these case we can't suppose that $(x,y,z,w)=1$

Audiophile wrote: Show that $(w, x, y, z)=(0,0,0,0)$ is the only integer solution to the equation $$w^{2}+11 x^{2}-8 y^{2}-12 y z-10 z^{2}=0$$ Equation is $w^2+11x^2=8(y-2z)^2+22z(2y-z)$ This implies $w^2\equiv 8(y-2z)^2\pmod{11}$ Since $8$ is not quadratic residue modulo $11$, this means $w\equiv y-2z\equiv 0\pmod{11}$ Setting $w=11w'$ and $y=2z+11u$, equation becomes $11^2w'^2+11x^2=8.11^2u^2+22z(3z+11u)$ And so $11w'^2+x^2=88u^2+6z^2+22zu$ And so $x^2\equiv 6z^2\pmod{11}$ Since $6$ is not quadratic residue modulo $11$, this means $x\equiv z\equiv 0\pmod{11}$ And so $y\equiv 0\pmod{11}$ And so $w\equiv x\equiv y\equiv z\pmod {11}$ And infinite descent modulo $11$ ends the proof.