The sum of all possible real value of $\text{'}a\text{'}$ for which vectors ${\overrightarrow{\mathrm{r}}}_1=\ln \left|\mathrm{a}\right|\widehat{\mathrm{i}}+\mathrm{a}\widehat{\mathrm{j}}-\widehat{\mathrm{k}}$ and ${\overrightarrow{\mathrm{r}}}_2=(1+{\mathrm{a}}^2)\widehat{\mathrm{i}}-\widehat{\mathrm{j}}-(\mathrm{a}-1)\widehat{\mathrm{k}}$  are orthogonal is :

${\overrightarrow{\mathrm{r}}}_1\cdot {\overrightarrow{\mathrm{r}}}_2=0$
 
$\begin{array}{l}\ln \left|\mathrm{a}\right|(1+{\mathrm{a}}^2)-\mathrm{a}+\mathrm{a}-1=0\\ \ln \left|\mathrm{a}\right|=\frac{1}{1+{\mathrm{a}}^2}\end{array}$
From the graph the two curves 

 $\mathrm{y}=\log \left|\mathrm{x}\right|,\mathrm{y}=\frac{1}{1+{\mathrm{x}}^2}$ intersect at two points ${\mathrm{x}}_1,{\mathrm{x}}_2$.

 it is clear that number of values of are two i.e., ${\mathrm{x}}_1,{\mathrm{x}}_2$ and ${\mathrm{x}}_1+{\mathrm{x}}_2=0$