If $2 x^2+y^2+8 z^2-2 \sqrt{2} x y+4 \sqrt{2} y z-8 z x=(A x+y+B z)^2$, then the value of $\left(A^2+B^2-A B\right)$ is:

14

(a – b – c)² = a² + b² + c² – 2ab + 2bc – 2ca

= 2x² + y² + 8z²– 2√2xy + 4√2yz – 8zx = (Ax + y + Bz)²

= (√2 x)² + y² + (2√2 z)² – 2 × (√2 x) × y + 2 × y × 2√2z – 2 × 2√2 z × √2 x = (Ax + y + Bz)²

= (√2x –y – 2√2 z)² = (Ax + y + Bz)²

Comparing on both sides

A = √2

B = – 2√2

Now,

(A² + B² – AB)

= [(√2)² + (–2√2)² – √2 × (–2√2)]

= [2 + 8 + 4] = 14