The distance between a point [tex](x,y,z)[/tex] on the given plane and the point (0, 2, 4) is
[tex]\sqrt{f(x,y,z)}=\sqrt{x^2+(y-2)^2+(z-4)^2}[/tex]
but since [tex]\sqrt{f(x,y,z)}[/tex] and [tex]f(x,y,z)[/tex] share critical points, we can instead consider the problem of optimizing [tex]f(x,y,z)[/tex] subject to [tex]x-2y+3z=6[/tex].
The Lagrangian is
[tex]L(x,y,z,\lambda)=x^2+(y-2)^2+(z-4)^2+\lambda(x-2y+3z-6)[/tex]
with partial derivatives (set equal to 0)
[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-4)+3\lambda=0\implies z=4-\dfrac{3\lambda}2[/tex]
[tex]L_\lambda=x-2y+3z-6=0\implies x-2y+3z=6[/tex]
Solve for [tex]\lambda[/tex]:
[tex]x-2y+3z=-\dfrac\lambda2-2(2+\lambda)+3\left(4-\dfrac{3\lambda}2\right)=6[/tex]
[tex]\implies2=7\lambda\implies\lambda=\dfrac27[/tex]
which gives the critical point
[tex]x=-\dfrac17,y=\dfrac{16}7,z=\dfrac{25}7[/tex]
We can confirm that this is a minimum by checking the Hessian matrix of [tex]f(x,y,z)[/tex]:
[tex]\mathbf H(x,y,z)=\begin{bmatrix}f_{xx}&f_{xy}&f_{xz}\\f_{yx}&f_{yy}&f_{yz}\\f_{zx}&f_{zy}&f_{zz}\end{bmatrix}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}[/tex]
[tex]\mathbf H[/tex] is positive definite (we see its determinant and the determinants of its leading principal minors are positive), which indicates that there is a minimum at this critical point.
At this point, we get a distance from (0, 2, 4) of
[tex]\sqrt{f\left(-\dfrac17,\dfrac{16}7,\dfrac{25}7\right)}=\sqrt{\dfrac27}[/tex]
