$\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}} \mathrm{and} \overrightarrow{\mathrm{c} } are three non zero non coplanar vectors and {\overrightarrow{\mathrm{r}}}_1=\mathrm{p}\overrightarrow{\mathrm{a}}+\mathrm{q}\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}} \mathrm{and} {\overrightarrow{\mathrm{r}}}_2=\overrightarrow{\mathrm{a}}+\mathrm{p}\overrightarrow{\mathrm{b}}+\mathrm{q}\overrightarrow{\mathrm{c}}$.  lf the vectors $\overrightarrow{{\mathrm{r}}_1}+2\overrightarrow{{\mathrm{r}}_2}\text{ and }2{\overrightarrow{\mathrm{r}}}_1+\overrightarrow{{\mathrm{r}}_2}$  are collinear, then (p, q) is

$\begin{array}{l}\overrightarrow{{\mathrm{r}}_1}+2{\overrightarrow{\mathrm{r}}}_2=(\mathrm{p}\overrightarrow{\mathrm{a}}+\mathrm{q}\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})+2(\overrightarrow{\mathrm{a}}+\mathrm{p}\overrightarrow{\mathrm{b}}+\mathrm{q}\overrightarrow{\mathrm{c}})\\ =(\mathrm{p}+2)\overrightarrow{\mathrm{a}}+(\mathrm{q}+2\mathrm{p})\overrightarrow{\mathrm{b}}+(1+2\mathrm{q})\overrightarrow{\mathrm{c}}\\ 2\overrightarrow{{\mathrm{r}}_1}+{\overrightarrow{\mathrm{r}}}_2=(2\mathrm{p}+1)\overrightarrow{\mathrm{a}}+(2\mathrm{q}+\mathrm{p})\overrightarrow{\mathrm{b}}+(2+\mathrm{q})\overrightarrow{\mathrm{c}}\\ \frac{\mathrm{p}+2}{2\mathrm{p}+1}=\frac{\mathrm{q}+2\mathrm{p}}{2\mathrm{q}+\mathrm{p}}=\frac{1+2\mathrm{q}}{2+\mathrm{q}}\\ =\frac{\mathrm{p}+\mathrm{q}+2\mathrm{p}+2\mathrm{q}+3}{\mathrm{p}+\mathrm{q}+2\mathrm{p}+2\mathrm{q}+3}=1\\ \Rightarrow \mathrm{p}=1\text{ and }\mathrm{q}=1\end{array}$