We consider the Diophantine inequalitywhere 1 < c < 15 14 , N is a sufficiently large real number and E > 0 is an arbitrarily large constant. We prove that the above inequality has a solution in primes p 1 , p 2 , p 3 such that each of the numbers p 1 + 2, p 2 + 2, p 3 + 2 has at most 369 180−168c prime factors, counted with the multiplicity.It follows from Theorem 1 that if c > 1 is close to 1, then inequality (1), with ∆ given by (3), has a solution in primes p i such that p i + 2 ∈ P 30 .We can establish a similar result for the inequality (1) with ∆ = N −κ for certain κ > 0, but then we would have p i + 2 ∈ P m , i = 1, 2, 3, where m depends on c and κ.Notations in the paper shall be as follows. By ε and A we denote an arbitrarily small positive number and respectively, an arbitrarily large constant which may not be the same in different formulae. The letter p always denotes a prime number. By τ (n), µ(n), ϕ(n) and Λ(n) we denote the number of divisors of n, Möbius' function, Euler's function and Von Mangoldt's function respectively. We shall use (m, n) and [m, n] for the greatest common divisor and the least common multiple of the integers m, n. (We denote in this way also open and closed intervals from the real line, but the meaning will be clear from the context). Let [t] be the integer part of the real number t and e(t) = e 2πit . With χ we denote a Dirichlet's character. As usual, χ (mod q) means that the summation is taken over all Dirichlet's characters modulo q. Respectively, χ (mod q) * means that the summation is taken over the primitive Dirichlet's characters modulo q.Suppose that χ is a Dirichlet's character and L(s, χ) is the corresponding L-function. If T ≥ 2 and 0 ≤ σ ≤ 1 we denote by N(T, σ, χ) the number of zeros ρ = β + iγ of L(s, χ) such that |γ| ≤ T and σ ≤ β ≤ 1. We also write N(T, χ) = N(T, 0, χ).