Let $(x_n, y_n, z_n) = (3, 1, 0)$ for $n=0$
For $n \ge 1$,
$$\begin{align} x_n &= x_{n-1} +3 z_{n-1}\\ y_n &= x_{n-1} +2 z_{n-1}\\ z_n &= 5 y_{n-1} \end{align}$$
Please let me know the formula to find $x_n,y_n,z_n$ values of any integer $n$.
The following would be the series:
$$\begin{array}{c:l} n & x_n,y_n,z_n\\ \hline 0 & 3,1,0\\ 1 & 3,3,5\\ 2 & 18,13,15\\ 3 & 63,48,65\\ \end{array}$$
On
$$\begin{align} x_n &= x_{n-1} +3 z_{n-1}\\ y_n &= x_{n-1} +2 z_{n-1}\\ z_n &= 5 y_{n-1} \end{align}$$ $$x_n-y_n=z_{n-1}=5y_{n-2}\implies x_n=y_n+5y_{n-2}$$ $$y_n=y_{n-1}+5y_{n-3}+2z_{n-1}=y_{n-1}+5y_{n-3}+10y_{n-2}$$ You'll get solving recurrence relation: $$y_n=f(n)$$ Then you'll get: $$z_n=5f(n-1)$$ and $$x_n=x_{n-1}+15f(n-2)$$ from where again solving a recurrence relation you'll get: $$x_n=g(n)$$
On
As pointed out by GerryMyerson, you can rewrite your problem as \begin{align} X_n=A\cdot X_{n-1} =\ldots=A^nX_0 \end{align} With $X_n =(x_n,y_n,z_n)^T$ and $X_0=(3,1,0)$, we only need the appropriate matrix $A$.
Your recurrence relation is \begin{align} x_n &=x_{n-1}+3z_{n-1}\\ y_n &=x_{n-1}+2z_{n-1}\\ z_n &= 5 y_{n-1} \end{align} Or as matrix-vector product \begin{align} \begin{pmatrix}x_{n-1} \\ y_{n-1}\\ z_{n-1} \end{pmatrix} = \underbrace{\begin{pmatrix}1 & 0 & 3\\ 1 & 0 & 2 \\ 0 & 5 &0 \end{pmatrix}}_{=:A} \begin{pmatrix}x_{n-1} \\y_{n-1}\\z_{n-1} \end{pmatrix} \end{align} Now we have to diagonalize $A$, i.e. finding matrices, such that $A=V D V^T$
Do you know how to proceed?
By finding the characteristic polynomial of the matrix $$ M = \left(\begin{matrix}1&0&3\\1&0&2\\0&5&0\end{matrix}\right) $$ that is $p(t)=t^3-t^2-10t-5$, and exploiting the Cayley-Hamilton theorem, we have that: $$ x_{n+3} = x_{n+2}+10 x_{n+1} + 5 x_n $$ and the same relation holds by replacing $x_{*}$ with $y_{*}$ or $z_{*}$. This gives an iterative method for finding $x_n,y_n,z_n$, and a closed expression too, in terms of linear combinations of the $n$-th powers of the roots of $p(t)$.