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Question
Find the value of p for which the following lines are perpendicular :
`(1-x)/3 = (2y-14)/(2p) = (z-3)/2 ; (1-x)/(3p) = (y-5)/1 = (6-z)/5`
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Solution
`(x-1)/-3 = (y-7)/p = (z-3)/2`
`& (x-1)/(-3p) = (y-5)/1 = (z-6)/-5`
are perpandicular
⇒ `(-3)(-3p) + (p)(1) + (2)(-5) = 0`
`9p + p -10 = 0`
`10 p = 10`
⇒ p = 1
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