1. Tardigrade
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  4. 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]

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Q. Let $\alpha, \beta, \gamma \in R$ and point $P(\alpha, \beta, \gamma)$ in $R^{3}$, consider three planes $P_{1}, P_{2}, P_{3}$ in $R^{3}$ where
$P_{1}: 2 x+3 y+6 z+30=0$
$P_{2}: 2 x+3 y+6 z+1=0$
$P_{3}: 2 x+3 y+6 z-5=0$
If $\frac{a}{b}$ is the probability that length of perpendicular from point $P$ to plane $P_{2}$ is less than equal to $1$ given that point $P$ lies between plane $P_{1}$ and $P_{3}$, then the value of $|3 a-b|$ is greater than
[Note : a, b are coprime numbers]

Probability - Part 2

Solution:

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