Maths

3, A? , A? , A? , .A? , 243
d = (243-3)/ (m+1) = 240/ (m+1)
3, G? , G? , G? , 243
r = (243/3)¹/ (³? ¹) = (81)¹/? = 3
G? = A?
=> 3 (3)² = 3 + 4 (240/ (m+1)
=> 27 = 3 + 960/ (m+1)
=> m+1 = 40
=> m=39

Normal of plane = |î? k|
|1 0|
|0 1 -1|
n = -î+? +k
D.R.'s = -1,1,1
Plane => -1 (x-1)+1 (y-0)+1 (z-0) = 0
=> x-y-z-1=0
If (x, y, z) is foot of perpendicular of M (1,0,1) on the plane then
(x-1)/-1 = (y-0)/1 = (z-1)/1 = - (1-0-1-1)/3
x=4/3, y=-1/3, z=2/3
α+β+γ = 4/3 - 1/3 + 2/3 = 5/3

x-2y+5z=0
-2x+4y+z=0
-7x+14y+9z=0
2x (i)+ (ii) => z=0
=> x=2y
=> 15 ≤ x²+y²+z² ≤ 150
=> 15 ≤ 4y²+y² ≤ 150
=> 3 ≤ y² ≤ 30
=> y = ±2, ±3, ±4, ±5
=> 8 solutions

Let xyz be the three digit number
x+y+z=10, x≥1, y≥0, z≥0
x-1=t => x=1+t, x-1≥0
t≥0
t+y+z = 10-1
t+y+z=9, 0≤t, y, z≤9
total number of non negative integral solution =? ³? ¹C? = ¹¹C? = (11*10)/2 = 55
But for t=9, x=10, so required number or integers = 55-1=54