Hello Nick,
Let's let our objective function be the square of the distance from the point on the plane $(x,y,z)$ to the given point $(0,1,1)$:
$$f(x,y,z)=x^2+(y-1)^2+(z-1)^2$$
The point $(x,y,z)$ is constrained to be on the given plane, hence our constraint is:
$$g(x,y,z)=x-2y+3z-6=0$$
Using Lagrange multipliers, we obtain the system:
$$2x=\lambda$$
$$2(y-1)=-2\lambda$$
$$2(z-1)=3\lambda$$
This implies:
$$\lambda=2x=1-y=\frac{2}{3}(z-1)$$
From this, we obtain:
$$y=1-2x,\,z=3x+1$$
Substituting for $y$ and $z$ into the constraint, we find:
$$x-2(1-2x)+3(3x+1)-6=0$$
Solving for $x$ (and using the values for $y$ and $z$ in terms of $x$) we find:
$$x=\frac{5}{14}\implies y=\frac{2}{7},\,z=\frac{29}{14}$$
Thus, the point on the plane $x-2y+3z=6$ that is closest to the point $(0,1,1)$ is:
$$\left(\frac{5}{14},\frac{2}{7},\frac{29}{14} \right)$$