$\Delta=\begin{bmatrix} 1 & log_x\;y & log_x\;z \\ log_y\;x & 1 & log_y\;z \\ log_z\;x & log_z\;y & 1 \end{bmatrix}$
From the information in the toolbox
$log _a\;b =\large\frac{log_e\;b}{log_e\;a}$
$\Delta=\begin{bmatrix} 1 & \large\frac{log\; y}{log\; x} & \large\frac{log\; z}{log\; x} \\ \large\frac{log \;x}{\log \;y} & 1 & \large\frac{log\; z}{log \;y} \\ \large\frac{log\;x}{log\; z} & \large\frac{log \;y}{log\; z} & 1 \end{bmatrix}$
Now apply $R_1 \to R_1.log\; x,R_2 \to R_2. log\; y, R_3 \to R_3.log \;z$
$\Delta=\large\frac{1}{log\; x\; log\; y\; log \;z}\begin{bmatrix} log\; x & log\; y & log\;z \\ log\; x & log \;y & log\; z \\ log\; x & log \;y & log\; z \end{bmatrix}$
But, since the rows are identical, the value of the determinant is zero
Therefore $\Delta =\large\frac{1}{log\;x\; log\; y\; log \;z} \times 0$
$\Delta =0$
Hence Proved.