The points P, Q, R and S on the sides OA, OB, OC and AB, respectively, of a regular tetrahedron OABC are   coplanar. It is  given  $\frac{\mathrm{OP}}{\mathrm{OA}}=\frac{1}{3},\frac{\mathrm{OQ}}{\mathrm{OB}}=\frac{1}{2},\frac{\mathrm{OR}}{\mathrm{OC}}=\frac{1}{3}\text{ and }\frac{\mathrm{OS}}{\mathrm{AB}}=\mathrm{\lambda } ,\mathrm{then} \mathrm{the} \mathrm{value} \mathrm{\lambda } \mathrm{is}$

$\begin{array}{l}\mathrm{Let} \overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{b}}\text{ and }\overrightarrow{\mathrm{OC}}=\overrightarrow{\mathrm{c}}\\ \mathrm{then} \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}\text{ and }\overrightarrow{\mathrm{OP}}=\frac{1}{3}\overrightarrow{\mathrm{a}}\\ \overrightarrow{\mathrm{OQ}}=\frac{1}{2}\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{OR}}=\frac{1}{3}\overrightarrow{\mathrm{c}}\end{array}$

Since P, Q,, R and S are coplanar, then $\overrightarrow{\mathrm{PS}}=\mathrm{\alpha }\overrightarrow{\mathrm{PQ}}+\mathrm{\beta }\overrightarrow{\mathrm{PR}}$

( $\overrightarrow{\mathrm{PS}}$ can be written as a linear  combination of  $\overrightarrow{\mathrm{PQ}}$and $\overrightarrow{\mathrm{PR}}$)

$\begin{array}{l}=\mathrm{\alpha }(\overrightarrow{\mathrm{OQ}}-\overrightarrow{\mathrm{OP}})+\mathrm{\beta }(\overrightarrow{\mathrm{OR}}-\overrightarrow{\mathrm{OP}})\\ \overrightarrow{\mathrm{OS}}-\overrightarrow{\mathrm{OP}}=-(\mathrm{\alpha }+\mathrm{\beta })\frac{\overrightarrow{\mathrm{a}}}{3}+\frac{\mathrm{\alpha }}{2}\overrightarrow{\mathrm{b}}+\frac{\mathrm{\beta }}{3}\overrightarrow{\mathrm{c}}\\ \Rightarrow \overrightarrow{\mathrm{OS}}=(1-\mathrm{\alpha }-\mathrm{\beta })\frac{\overrightarrow{\mathrm{a}}}{3}+\frac{\mathrm{\alpha }}{2}\overrightarrow{\mathrm{b}}+\frac{\mathrm{\beta }}{3}\overrightarrow{\mathrm{c}}---\left(\mathrm{i}\right)\\ \mathrm{given} \overrightarrow{\mathrm{OS}}=\mathrm{\lambda }\overrightarrow{\mathrm{AB}}=\mathrm{\lambda }(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}})---\left(\mathrm{ii}\right)\end{array}$

$\text{ From (i) and (ii), }\beta =0,\frac{1-\alpha }{3}=-\lambda \text{ and }\frac{\alpha }{2}=\lambda$

$\begin{array}{r}\Rightarrow 2\lambda =1+3\lambda \\ \text{ or }\lambda =-1\end{array}$