Q:

On the question regarding when the determinant's answer is going to be prime?

Let $a$ and $b$ are two consecutive prime numbers. 
$$\begin{vmatrix}
3 & b \\ 
2 & a 
\end{vmatrix}$$
Gauss's Prime number theorem is the best object to study about prime numbers between any two natural numbers. But this theorem can also be used to have a matrix whose determinant is always a prime number.
Let $p$ be any prime number,
Using Prime number theorem we have,
$$f(x) ={ x \over \ln(x)}$$
Now, 
$$\pi(p-1) - \pi(3) = A$$
Here, A = even number.
Now, determinant is as follows:
$$\begin{vmatrix}
2 & -1 & 3 - p \\ 
3 & A  & 2\\ 
p & -3  & p-1
\end{vmatrix} \pm 1$$
The value of above Determinant is always a prime number.
I have doubt that most probably, my logic in the 3x3 determinant is true, but in 2x2 determinant, I want to know that if a and b are two consecutive primes when the answer of 2x2 and 3x3 determinant will be a prime? 

A:

Let's consider the determinant 
$$\begin{vmatrix}
3 & b \\ 
2 & a 
\end{vmatrix}$$
Which is actually $3a-2b$ and it's not well controlled especially when our demand is of output which should be prime number. 
But I have something which might interest you in some way. 
Once upon a time I Conjectured that $3p_n + 2p_{n-1}$ is either prime or semiprime, but it's validity was transient. But for sure it does take Infinitude of primes. Note that $p_n$ is the $nth$ prime. 

I'll assume that you meant exact count of primes, not asymptotically. 
Let's define $T(x)$ to be the exact count of prime numbers below any given number $x$ 
Now considering your following determinant
$$\begin{vmatrix}
2 & -1 & 3 - p \\ 
3 & A  & 2\\ 
p & -3  & p-1
\end{vmatrix} \pm 1$$ 
Where $A$ is meant to be an positive even integer such that $T(p-1) - T(3) = A$ 
Your claim is such that the value of determinant is always a prime. 
The determinant boils down to 
$Ap^2 + p(A+10) -1(30+2A) \pm1$ 
By solving the above Quadratic in any prime $p$ we can obtain the values of $p$ which are admissible. 
For negative case, which is 
$Ap^2 + p(A+10) -(31+2A)$ we can easily prove that there doesn't exist any such prime $p$, while assuming everything same as you mentioned. 
Counterexample
Consider $p=19$ we have $T(19-1) - 1 = A = 6$ 
Plug in $p=19$ and $A=6$ in 
$Ap^2 + p(A+10) -1(30+2A) \pm1$ we get $2428 \pm1$ which yields $2427$ and $2429$ none of them are primes. 

