Very good result Dibakar Pal, congratulation

Dear Risto,

Your question is nice and deep.In fact, it is related to the gap between prime numbers.  

As a partial result, if k = 1, the only solution of the system is 3, 5 and 7. If k =2,  (3, 7, 11)  is a solution and if k = 3, then (7, 13,19 ) is a solution. Also If k = 4, ( 3, 11 and 19) and for k =5  ( 3,13,23).  For k = 6, ( 5, 17,29) is a solution, etc...,  The problem reduced into 

Is the triple  ( p, p+2k, p+4k  ) include primes for some positive value of k and prime p? I conjecture that your claim is true. 

 Best wishes

As observed by Issam Kaddoura, the problem by Professsor Malčeski consists in the fact that for a given positive integer k to find a prime p such that p, p + 2k and p + 4k are all primes. If the previously mentioned condition „...to find a prime p...“ we replace by the condition „...to find an integer p...“, then the related problem becomes a paricular case (with k = 3) of Dickson's conjecture stated by Dickson (Dickson, L. E. (1904), "A new extension of Dirichlet's theorem on prime numbers", Messenger of mathematics, vol. 33, pp. 155–161). Namely, Dickson's conjecture asserts that if for a finite set of k linear forms a1 + b1n, a2 + b2n, ..., ak + bkn (ai and bi are integers and bi ≥ 1 for all i = 1, 2,..., k ), there does not exist any integer d dividing all the products (a1 + b1n, )(a2 + b2n) ...(ak + bkn) for every integer n, then there exist infinitely many positive integers n for which all numbers a1 + b1n, a2 + b2n, ..., ak + bkn are primes.

Nevertheless the fact that the question propsed by Professor Malčeski is in relation to the stronger particular 3-version of Dickson's conjecture, I believe that computational results would be suggested the positive answer to this question.