Let α, β, γ ∈ R and point P(α, β, γ) in R3, consider three planes P1, P2, P3 in R3 where P1: 2 x+3 y+6 z+30=0 P2: 2 x+3 y+6 z+1=0 P3: 2 x+3 y+6 z-5=0 If (a/b) is the probability that length of perpendicular from point P to plane P2 is less than equal to 1 given that point P lies between plane P1 and P3, then the value of |3 a-b| is greater than [Note: a, b are coprime numbers]

Question

Let α, β, &gamma; ∈ R and point P(α, β, &gamma;) in R3, consider three planes P1, P2, P3 in R3 where P1: 2 x+3 y+6 z+30=0 P2: 2 x+3 y+6 z+1=0 P3: 2 x+3 y+6 z-5=0 If (a/b) is the probability that length of perpendicular from point P to plane P2 is less than equal to 1 given that point P lies between plane P1 and P3, then the value of |3 a-b| is greater than [Note: a, b are coprime numbers] (A) 2 (B) 3 (C) 4 (D) 5

Tardigrade Admin · Accepted Answer

Distance between P2 P3 is (6/7) P1 P3 is 5, P2 P1 is (29/7) ∴ Probability =(1+(6/7)/5)=(13/35)

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Distance between P2​ & P3​ is 76​ P1​ & P3​ is 5,P2​ & P1​ is 729​ ∴ Probability =51+76​​=3513​

Distance between $P_{2}$ & $P_{3}$ is $\frac{6}{7}$
$P_{1}$ & $P_{3}$ is $5, P_{2}$ & $P_{1}$ is $\frac{29}{7}$
$∴$ Probability $= \frac{1 + \frac{6}{7}}{5} = \frac{13}{35}$