The roots of the equation $(\mathrm{x}-\mathrm{a})(\mathrm{x}-\mathrm{b})+(\mathrm{x}-\mathrm{b})(\mathrm{x}-\mathrm{c})+(\mathrm{x}-\mathrm{c})(\mathrm{x}-\mathrm{a})=0$ are real and equal if

The given equation is $(\mathrm{x}-\mathrm{a})(\mathrm{x}-\mathrm{b})+(\mathrm{x}-\mathrm{b})(\mathrm{x}-\mathrm{c})+(\mathrm{x}-\mathrm{c})(\mathrm{x}-\mathrm{a})=0$

This can be rewrite as $3x^2-2x\left(a+b+c\right)+ab+bc+ca=0$

Given that the roots of the above equaiton are real and equal, so that the discriminant is equal to zero

It implies that $b^2-4ac=0$

Hence, 

$\begin{array}{rcl}4{\left(a+b+c\right)}^2-4\left(3\right)\left(ab+bc+ca\right)& =& 0\\ a^2+b^2+c^2-ab-bc-ca& =& 0\\ 2a^2+2b^2+2c^2-2ab-2bc-2ca& =& 0\\ {\left(a-b\right)}^2+{\left(b-c\right)}^2+{\left(c-a\right)}^2& =& 0\end{array}$

Sum of three square terms is zero implies that each term must be equal to zero

it implies that $a-b=0,b-c=0,c-a=0$

Therefore, $\boxed{a=b=c}$