If x→,y→  are two non-zero and non-collinear vectors satisfying  (a−2)α2+(b−3)α+cx→+(a−2)β2+(b−3)β+cy→+(a−2)γ2+(b−3)γ+c(x→×y→)=0 where α,β,γ are three distinct real numbers, then find the value of a2+b2+c2−4

Since x→ and y→ are non-collinear vectors, therefore x→,y→ and x→×y→ are non-coplanar vectors. (a−2)α2+(b−3)α+c+(a−2)β2+(b−3)ββ+c]y→+(a−2)γ2+(b−3)γ+c(x→×y→)=0 Coefficient of each vector x→,y→ and x→×y→ is zero. (a−2)α2+(b−3)α+c=0(a−2)β2+(b−3)β+c=0(a−2)γ2+(b−3)β+c=0The above three equations will satisfy if the coefficients of α,β and  γ are zero because α,β and γ are three distinct real numbers .a−2=0 or a=2b−3=0 or b=3 and c=0∴ a2+b2+c2=22+32+02=4+9=13