If $\overrightarrow{\mathrm{P}}=\overrightarrow{\mathrm{i}}+\mathrm{\lambda }\overrightarrow{\mathrm{j}}+\overrightarrow{\mathrm{k}}, \overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{i}}+\overrightarrow{\mathrm{j}}+\overrightarrow{\mathrm{k}}$ and $\left|\overrightarrow{\mathrm{P}}+\overrightarrow{\mathrm{q}}\right|=\left|\overrightarrow{\mathrm{P}}\right|+\left|\overrightarrow{\mathrm{q}}\right|$ then $\mathrm{\lambda }=$

$$\begin{gathered} \overline{\mathrm{p}}+\overline{\mathrm{q}}=2\mathrm{i}+(\mathrm{\lambda }+1)\mathrm{j}+2\mathrm{k} \\ \left|\overline{\mathrm{p}}+\overline{\mathrm{q}}\right|=\left|\overline{\mathrm{p}}\right|+\left|\overline{\mathrm{q}}\right| \\ \sqrt{4+{(\mathrm{\lambda }+1)}^2+4}=\sqrt{1+{\mathrm{\lambda }}^2+1}+\sqrt{1+1+1} \\ \mathrm{S}.\mathrm{B}.\mathrm{S} \\ {\mathrm{\lambda }}^2+2\mathrm{\lambda }+9={\mathrm{\lambda }}^2+2+3+2\sqrt{3} \sqrt{{\mathrm{\lambda }}^2+2} \\ 2\mathrm{\lambda }+4=2\sqrt{3} \sqrt{{\mathrm{\lambda }}^2+2} \\ \mathrm{S}.\mathrm{B}.\mathrm{S} \\ {\mathrm{x}}^{2}4+4\mathrm{\lambda }=3{\mathrm{\lambda }}^2+6 \\ \Rightarrow 2{\mathrm{\lambda }}^2-4\mathrm{\lambda }+2=0 \Rightarrow {\mathrm{\lambda }}^2-2\mathrm{\lambda }+1=0 \\ \Rightarrow \mathrm{\lambda }=1 \end{gathered}$$