show that (1,1,1) is the incentre of the tetrahedron formed by four planes x=0,y=0,z=0,x+2y+2z-2=0

Dear Student

Clearly, the planes x = 0, y = 0 and z = 0 meet in (0, 0, 0)
Hence, the incentre lies on the perpendicular from (0, 0, 0) to the plane x+2y+2z-2=0and divides it in the ratio 3 : 1, i.e. 3 from the vertex (0, 0, 0) and 1 from the plane x+2y+2z-2=0
The equation of the perpendicular from (0, 0, 0) to the plane x+2y+2z-2=0 is x/1 = 2y = 2z = λ(say)
Any point on the perpendicular is (λ, λ/2,λ/2)
If it lies on the planex+2y+2z-2=0
so, λ= 2/3
Thus, the perpendicular from (0, 0, 0) meets the planex+2y+2z-2=0at ( 2/3 , 1/3 , 1/3)

Also incentre divides line of join of (0,0,0 ) and( 2/3 , 1/3 , 1/3)in 3:1
Let , (a,b,c) be incentre
a= 3.(2/3)/4 = ½
b = 3.(1/3) / 4 = ¼
c = 1/4
Please check the question as the results is not coming as per question to prove

Thanks