If the distance of a point lying on the plane 2 x + 3 y + 6 z = p from the point ( 3 , 0 , 1 ) is unity then the total number of possible values of p , where p is a prime number, is

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2 2 + 3 2 + 6 2 ​ ∣2 ( 3 ) + 3 + 6 ( 1 ) − p ∣ ​ ≤ 1 ⇒ ∣12 − p ∣ ≤ 7 ⇒ − 7 ≤ p − 12 ≤ 7 ⇒ 5 ≤ p ≤ 19 ⇒ 5 , 7 , 11 , 13 , 17 , 19 i.e. six possible values of p .

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