Use lagrange multipliers to find the point on the plane x − 2y + 3z = 6 that is closest to the point (0, 2, 5).

Lagrangian:

[tex]L(x,y,z,\lambda)=x^2+(y-2)^2+(z-5)^2+\lambda(x-2y+3z-6)[/tex]

where the function we want to minimize is actually [tex]\sqrt{x^2+(y-2)^2+(z-5)^2}[/tex], but it's easy to see that [tex]\sqrt{f(\mathbf x)}[/tex] and [tex]f(\mathbf x)[/tex] have critical points at the same vector [tex]\mathbf x[/tex].

Derivatives of the Lagrangian set equal to zero:

[tex]L_x=2x+\lambda=0\implies x=-\dfrac\lambda2[/tex]
[tex]L_y=2(y-2)-2\lambda=0\implies y=2+\lambda[/tex]
[tex]L_z=2(z-5)+3\lambda=0\implies z=5-\dfrac{3\lambda}2[/tex]
[tex]L_\lambda=x-2y+3z-6=0[/tex]

Substituting the first three equations into the fourth gives

[tex]-\dfrac\lambda2-2(2+\lambda)+3\left(5-\dfrac{3\lambda}2\right)=6[/tex]
[tex]11-7\lambda=6\implies \lambda=\dfrac57[/tex]

Solving for [tex]x,y,z[/tex], we get a single critical point at [tex]\left(-\dfrac5{14},\dfrac{19}7,\dfrac{55}{14}\right)[/tex], which in turn gives the least distance between the plane and (0, 2, 5) of [tex]\dfrac5{\sqrt{14}}[/tex].