*Hint: In order to find the distance between the two points the formula would be √((x2 - x1)^2 + (y2 - y1)^2)
Now that we know the formula, all we do is plug in the numbers.
√((2 - 2)^2 + (2.5 - 1.2)^2)
√ 0 + (1.3)^2
√1.69
1.3
The distance between the two points is 1.3.
What I don’t speak that language im sorry
I assume there are some plus signs that aren't rendering for some reason, so that the plane should be

.
You're minimizing

subject to the constraint

. Note that

and

attain their extrema at the same values of

, so we'll be working with the squared distance to avoid working out some slightly more complicated partial derivatives later.
The Lagrangian is

Take your partial derivatives and set them equal to 0:

Adding the first three equations together yields

and plugging this into the first three equations, you find a critical point at

.
The squared distance is then

, which means the shortest distance must be

.