1. Write the points on the three lines in coordinate form

Given:

• $L_1: \vec r = \lambda \hat i$
  So a general point on $L_1$ is $P=(\lambda,0,0).$

• $L_2: \vec r = \hat k + \mu \hat j$
  So a general point on $L_2$ is $Q=(0,\mu,1).$

• $L_3: \vec r = \hat i+\hat j+v\hat k$
  So a general point on $L_3$ is $R=(1,1,v).$

We need those points $Q$ on $L_2$ for which there exist some $P\in L_1$ and $R\in L_3$ such that $P,Q,R$ are collinear.

2. Use collinearity condition

If $P,Q,R$ are collinear, then vectors $\overrightarrow{PQ}$ and $\overrightarrow{PR}$ must be parallel.

Now, $\overrightarrow{PQ}=Q-P=( -\lambda,\mu,1),$ $\overrightarrow{PR}=R-P=(1-\lambda,1,v).$

For collinearity, there must exist some scalar $t$ such that $(-\lambda,\mu,1)=t(1-\lambda,1,v).$

Equating components: $-\lambda=t(1-\lambda) \quad ...(1)$ $\mu=t \quad ...(2)$ $1=tv \quad ...(3)$

We need only the possible values of $\mu$.

3. Find condition on $\mu$ using two-point form of line

A cleaner way is to parametrize the line joining $P=(\lambda,0,0)$ and $R=(1,1,v)$.

A general point on line $PR$ is $P+s(R-P).$ So its coordinates are $(x,y,z)=(\lambda,0,0)+s(1-\lambda,1,v).$

Thus, $x=\lambda+s(1-\lambda), \quad y=s, \quad z=sv.$

Since $Q$ lies on $L_2$, we need $Q=(0,\mu,1).$ Hence, $y=\mu=s,$ so $s=\mu.$ Also, $z=1=sv=\mu v,$ so as long as $\mu\neq 0$, we can choose $v=\frac{1}{\mu}.$ Now impose the $x$-coordinate condition: $0=\lambda+\mu(1-\lambda).$ Simplify: $0=\lambda+\mu-\mu\lambda,$ $\lambda(1-\mu)=-\mu.$ So, $\lambda=\frac{-\mu}{1-\mu}, \qquad (\mu\neq 1).$

Therefore, for any $\mu\neq 0,1$, we can choose suitable $\lambda$ and $v$.

What about $\mu=1$? Then from $0=\lambda+1(1-\lambda)=1,$ which is impossible. So $\mu=1$ is not allowed.

What about $\mu=0$? Then $Q=(0,0,1)$ and from $1=\mu v$ impossible. So $\mu=0$ is also not allowed.

Hence the required points on $L_2$ are exactly those with $\mu\neq 0,1.$

4. Check the options

On $L_2$, point is $Q=(0,\mu,1)=\hat k+\mu \hat j$.

• A: $\hat k$ corresponds to $\mu=0$
  Not possible.

• B: $\hat k+\hat j$ corresponds to $\mu=1$
  Not possible.

• C: $\hat k+\frac12\hat j$ corresponds to $\mu=\frac12$
  Possible.

• D: $\hat k-\frac12\hat j$ corresponds to $\mu=-\frac12$
  Possible.

5. Final answer

The correct options are $\boxed{C, D}.$