Let $R=\begin{bmatrix} a & 3 & b \\ c & 2 & d \\ 0 & 5 & 0 \end{bmatrix}$ : a, b, c, d $\in \{0, 3, 5, 7, 11…

We know that, for invertible matrices $\left|R\right|\ne 0$

Now $\left|R\right|$ can be zero in following cases:

(i) Two of $a, b, c, d$ are zeroes which can be ($a$ and $b$),($b$ and $d$),($d$ and $c$) or ($c$ and $a$)

$\to 4\times 7^2\text{ ways }=196$

(ii) Any three of $a, b, c, d$ are zeroes

$\to {}^{4}C_3\times 7=28$

(iii) All four of $a, b, c, d$ are zeroes

$\to 1$

(iv) All four of $a, b, c, d$ are non-zero but same number

$\to 7$

(v) When two are alike and $2$ other are alike (non-zero)

$\to 7{\mathrm{C}}_2\times 2\times 2=84$

Number of invertible matrices

$=8^4-196-28-1-7-84=3780$