Guse lagrange multipliers to find the point on the plane x − 2y + 3z = 6

Let's let our objective function be the square of the distance from the point on the plane  (x,y,z)(x,y,z)  to the given point  (0,1,1)(0,1,1): f(x,y,z)=x2+(y−1)2+(z−1)2f(x,y,z)=x2+(y−1)2+(z−1)2the point  (x,y,z)(x,y,z)  is constrained to be on the given plane, hence our constraint is: g(x,y,z)=x−2y+3z−6=0g(x,y,z)=x−2y+3z−6=0using lagrange multipliers, we obtain the system: 2x=λ2x=λ2(y−1)=−2λ2(y−1)=−2λ2(z−1)=3λ2(z−1)=3λthis implies: λ=2x=1−y=23(z−1)λ=2x=1−y=23(z−1)from this, we obtain: y=1−2x,z=3x+1y=1−2x,z=3x+1substituting for  yy  and  zz  into the constraint, we find: x−2(1−2x)+3(3x+1)−6=0x−2(1−2x)+3(3x+1)−6=0solving for  xx  (and using the values for  yy  and  zz  in terms of  xx) we find: x=514⟹y=27,z=2914x=514⟹y=27,z=2914thus, the point on the plane  x−2y+3z=6x−2y+3z=6  that is closest to the point  (0,1,1)(0,1,1)  is: (514,27,2914)

Your answer should be “c” mark me brainliest

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