Using properties of determinants, prove that $\left|\begin{array}{ccc}1& 1& 1+3\mathrm{x}\\ 1+3\mathrm{y}& 1& 1\\ 1& 1+3\mathrm{z}& 1\end{array}\right|=9(3\mathrm{xy}+\mathrm{xy}+\mathrm{yz}+\mathrm{zx})$

From the question assume,
$\text{ L.H.S. }=\left|\begin{array}{ccc}1& 1& 1+3\mathrm{x}\\ 1+3\mathrm{y}& 1& 1\\ 1& 1+3\mathrm{z}& 1\end{array}\right|$
Taking x, y and z common from R₁, R₂ and R₃ respectively, we get
$=\mathrm{xyz}\left|\begin{array}{ccc}\frac{1}{\mathrm{x}}& \frac{1}{\mathrm{x}}& \frac{1}{\mathrm{x}}+3\\ \frac{1}{\mathrm{y}}+3& \frac{1}{\mathrm{y}}& \frac{1}{\mathrm{y}}\\ \frac{1}{\mathrm{z}}& \frac{1}{\mathrm{z}}+3& \frac{1}{\mathrm{z}}\end{array}\right|$
On applying R₁$\to$R₁+R+₂+R₃, we get
$=\mathrm{xyz}\left|\begin{array}{ccc}\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{z}}+3& \frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{z}}+3& \frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{z}}+3\\ \frac{1}{\mathrm{y}}+3& \frac{1}{\mathrm{y}}& \frac{1}{\mathrm{y}}\\ \frac{1}{\mathrm{z}}& \frac{1}{\mathrm{z}}+3& \frac{1}{\mathrm{z}}\end{array}\right|$
Taking common $\left(\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{z}}+3\right)$from R₁, we get
$=\mathrm{xyz}\left(\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{z}}+3\right)\left|\begin{array}{ccc}1& 1& 1\\ \frac{1}{\mathrm{y}}+3& \frac{1}{\mathrm{y}}& \frac{1}{\mathrm{y}}\\ \frac{1}{\mathrm{z}}& \frac{1}{\mathrm{z}}+3& \frac{1}{\mathrm{z}}\end{array}\right|$
On applying ${\mathrm{C}}_2\to {\mathrm{C}}_2-{\mathrm{C}}_1$and ${\mathrm{C}}_3\to {\mathrm{C}}_3-{\mathrm{C}}_1$, we get
$=\mathrm{xyz}\left(\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{z}}+3\right)\left|\begin{array}{ccc}1& 0& 0\\ \frac{1}{y}+3& -3& -3\\ \frac{1}{z}& 3& 0\end{array}\right|$
On expanding along R₁, we get
$\begin{array}{l}=\mathrm{xyz}\left(\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{z}}+3\right)1(0+9)\\ =\mathrm{xy}z\left(\frac{\mathrm{yz}+\mathrm{xz}+\mathrm{xy}+3\mathrm{xyz}}{\mathrm{xyz}}\right)\left(9\right)\\ =9(3\mathrm{xyz}+\mathrm{xy}+\mathrm{yz}+\mathrm{zx})\end{array}$
=RHS
Therefore, LHS=RHS has proven.