Three quadratic equations are given:
I. (x2 - 12x + 35 = 0)
II. (y2 + 3y - 70 = 0)
III. (z2 - 5z - 14 = 0) Three conditions are also given as A, B and C, which have to be considered for the roots of each equation: A: (x = y >
z)
B: (x < y >
z)
C: (x <
y)

I: x² - 12x + 35 = 0 or, (x - 5)(x - 7) = 0
so, x = 5,7 II: y² + 3y - 70 = 0 or, (y + 10)(y - 7) = 0
So, y = -10, 7 III: (z² - 5z - 14 = 0   or, (z - 7)(z + 2) = 0
So, z = 7,-2) 2) Condition A: (x = y > z)
Only possible (x = y = 7), then (z) must be less than 7 → (z = -2)
So ((x, y, z) = (7,7,-2))
Then (y + z = 7 + (-2) = 5 < x =7) True Condition B: (xz)
(y) must be 7 (cannot be -10).
Then (x < 7 or, x = 5), and (z < 7 or, z = -2)
So ((x, y, z) = (5, 7,-2))
Then (y + z = 5 < x = 5) False Condition C: (x
Again (y = 7), (x = 5).
(z) can be (7) or (-2). If (z = -2): (y + z = 5 < x = 5) False If (z = 7): (y + z = 14 < x = 5) False
So never true under C. ((y + z) < x) is true only when condition A follows. Hence, option a is correct.