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Suppose that `vec p, vec q and vec r` are three non-coplanar vectors in `R^3`. Let the components of a vector `vec s` along `vec p, vec q and vec r` be 4, 3 and 5, respectively. If the components of this vector `vec s` along `(-vec p+vec q +vec r), (vec p-vec q+vec r) and (-vec p-vec q+vec r)` are x, y and z, respectively, then the value of `2vec x+vec y+vec z` is

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The correct Answer is:
`(9)`
Here `s= 4P + 3q +5r`
and `s= (-P+q+r )x+ (p-q+r) y+(-P-q+r) z ……(1)`
` :. 4p+3q+5r =p (-x+y-z) +q(x-y+z)+ r(x+y+z)`
On comparing both sides we get
`-x+y -z= 4, x -y =3 " and " x+y+z=6`
On solving above equations we get
`x= 4 , y= (9)/(2), =(-7)/(2)`
`:. 2x +y+ z=8 + (9)/(2)- (7)/(2) = 9`
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