If ` vec x , vec y` are two non-zero and non-collinear vectors satisfying `[(a-2)alpha^2+(b-3)alpha+c] vec x+[(a-2)beta^2+(b-3)beta+c] vec y+[(a-2)gamma^2+(b-3)gamma+c]( vec xxx vec y)=0,w h e r ealpha,beta,gamma` are three distinct real numbers, then find the value of `(a^2+b^2+c^2-4)dot`

Answer

9

Explanation

Since `vecx and vecy` are non- collinear vectors, therefore, `vecx, vecy and vec xx vecy` are non- coplanar vectors.
`[(veca -2)alpha^(2)+(b+3)alpha+c]+[(a-2)beta^(2)+ (b-3)betabeta+c] vecy +[(a -2)gamma^(2)+(b-3)gamma+c] (vecx xx vecy)=0`
Cofficient of each vector `vecx, vecy and vecx xx vecy` is zero.
`(a-2)alpha^(2)+(b-3)alpha+c=0`
`(a-2)beta^(2)+(b-3)beta+c=0`
` (a-2)gamma^(2)+(b-3)beta+c=0`
The above three equations will satisfiy if the coefficients of `alpha,beta and gamma` are zero because `alpha, beta and gamma` are three distinct real numbers
`a-2=0 or a =2`
`b-3=0 or b=3 and c=0`
`a^(2)+b^(2) +c^(2) = 2^(2)+3^(2)+0^(2)=4+9=13`