$Three lines$ $L_1:\overrightarrow{r}=\lambda \widehat{i},\lambda \in ℝ$, $L_2:\overrightarrow{r}=\widehat{k}+\mu \widehat{j},\mu \in ℝ$, $L_3:\overrightarrow{r}=\widehat{i}+\widehat{j}+\nu \widehat{k},\nu \in ℝ$  $are given. For which the points$$Q$ $on$ $L_2$  $can we find$ $P$ $on$ $L_1$$and a point R on L_3 so that P, Q,R are collinear$

$\text{Position vector of point }P is \overrightarrow{p}=\lambda \hat{i}$

$\text{Position vector of }Q is \hat{q}=\mu \hat{j}+\hat{k}$

$Position vector of \text{point }R is \overrightarrow{r}=\hat{i}+\hat{j}+r\hat{k}$

$\mathrm{PQR}\text{ are collinear. Hence }\mathrm{x}\left(\overrightarrow{\mathrm{PQ}}\right)=\mathrm{y}\left(\overrightarrow{\mathrm{PR}}\right)$

$$\begin{gathered} \Rightarrow x\left(-\lambda i+\mu j+k\right)=y\left(\left(1-\lambda \right)i+j+rk\right) \\ It gives \frac{x}{y}=\frac{1-\lambda }{-\lambda }=\frac{1}{\mu }=r \end{gathered}$$

$\Rightarrow \overrightarrow{\mathrm{q}}=\frac{1}{\mathrm{r}}\hat{\mathrm{j}}+\hat{\mathrm{k}}\text{ or }\overrightarrow{\mathrm{q}}=\frac{\lambda }{\lambda -1}\hat{\mathrm{j}}+\hat{\mathrm{k}},\text{ where }\mathrm{r}\ne 0,\lambda \ne 0,\frac{\lambda }{\lambda -1}\ne 1$

$\Rightarrow \mu \ne 0,1$

$\text{ Hence, }\overrightarrow{\mathrm{q}}\ne \hat{\mathrm{k}}\text{ or }\hat{\mathrm{j}}+\hat{\mathrm{k}}$