A triangle has sides having equations equal to 2+2y-5=0, 2x-y+2=0 and 2x-y-10=0. Find the equation of the circle inscribed in the triangle.?

I'm glad you included a picture, because that will help us. You copied the problem down incorrectly when you wrote 2x - y + 2 = 0, when it should be 2x + y + 2 = 0

If you graph the lines, you'll see where the circle needs to be centered

A has a center at (2 , 1)
B has a center at (1 , -1)
C has a center at (-1 , 1)
D has a center at (1 , 1)

You'll see that you have an isosceles triangle with an axis of symmetry at x = 2.  That leaves option A for us.  The rest just won't work.  And the best part here is that when you graph it, even A doesn't work.

There's just all sorts of issues with this problem as it is presented.

The second line (2x - y + 2 = 0) and the third line (2x - y - 10 = 0) are parallel. Hence, no triangle is formed.

Corrected equations: 2 + 2y – 5 = 0, 2x + y + 2 = 0, 2x – y – 10 = 0
They make the triangle shown here
https://www.wolframalpha.com/input/?i=2+%2B+2y+%E2%80%93+5+%3D+0%2C+2x+%2B+y+%2B+2+%3D+0%2C+2x+%E2%80%93+y+%E2%80%93+10+%3D+0%2C+%28x+from+-2+to+6%29%2C+%28y+from+-6+to+1.6%29

The centre of the circle is on the angle bisectors of this triangle.
A bisector is the set of points which have the same distance from the edges of the angle. This link shows calculations for a similar example and details how to find the equations for two of the angle bisector lines. You do that.

https://study.com/academy/answer/the-sides-of-a-triangle-are-on-the-lines-6x-7y-plus-11-0-2x-plus-9y-plus-11-0-and-9x-2y-11-0-find-the-equation-of-the-circle-inscribed-in-the-triangle.html

The centre is on the intersection of those two lines, at a point P(a, b) say.
Since one of the triangle lines is y = 3/2 the radius will be r =  3/2 – b.

You could check your calculations by graphing the circle and line equations together, (using WolframAlpha) and confirming that the lines forming the triangle work out to be tangent to the inscribed circle.