$\mathrm{a},\mathrm{b},\mathrm{c}\text{ are distinct real numbers, not equal to one. If }\mathrm{ax}+\mathrm{y}+\mathrm{z}=0,\mathrm{x}+\mathrm{by}+\mathrm{z}=0\text{, and }$$\mathrm{x}+\mathrm{y}+\mathrm{cz}=0\text{ have a non trivial solution, then the value of }\frac{1}{1-\mathrm{a}}+\frac{1}{1-\mathrm{b}}+\frac{1}{1-\mathrm{c}}\text{ is equal }$to

Since the system has non-trivial solution, we have
$\left|\begin{array}{l}\mathrm{a}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{b}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{c}\end{array}\right|=0$
$\text{ Applying }{\mathrm{R}}_1\to {\mathrm{R}}_1-{\mathrm{R}}_2,{\mathrm{R}}_2\to {\mathrm{R}}_2-{\mathrm{R}}_3\text{, we get }$
$\mathrm{\Delta }=\left|\begin{array}{ccc}\mathrm{a}-1& 1-\mathrm{b}& 0\\ 0& \mathrm{b}-1& 1-\mathrm{c}\\ 1& 1& \mathrm{c}\end{array}\right|=0$
Or $\mathrm{c}(1-\mathrm{a})(1-\mathrm{b})+(1-\mathrm{b})(1-\mathrm{c})-(1-\mathrm{c})(\mathrm{a}-1)=0$
$\text{ Dividing throughout by }(1-\mathrm{a})(1-\mathrm{b})(1-\mathrm{c})\text{, we get }$
$\begin{array}{l}\frac{\mathrm{c}}{1-\mathrm{c}}+\frac{1}{1-\mathrm{c}}+\frac{1}{1-\mathrm{b}}=0\\ \text{ Or }-1+\frac{1}{1-\mathrm{c}}+\frac{1}{1-\mathrm{b}}+\frac{1}{1-\mathrm{a}}=0\\ \frac{1}{1-\mathrm{c}}+\frac{1}{1-\mathrm{a}}+\frac{1}{1-\mathrm{b}}=1\end{array}$