Here is one way of doing this.

The point (1,k) is anywhere on the vertical line x=1.

The distance of vertical line (x=1) from the vertical line (x = -2) is 3 units.

The distance of (1,k) from (2,0) can be found by the distance formula.

So,

3 = sqrt[(1-2)^2 +((k-0)^2]

3 = sqrt[1 +k^2]

Square both sides.

9 = 1 +k^2

k^2 = 9 -1 = 8

k = +,-sqrt(8) = +,-2sqrt(2) --------answer.

-------------

Check if (1, +,-2sqrt(2)) is 3 units away from (2,0):

d = sqrt[(1 -2)^2 +(+,-2sqrt(2) -0)^2]

d = sqrt[(1) +4(2)]

d = sqrt(9) = 3 ------it is, so, okay.