Let $L_1:\vec{r}=(\hat{i}-\hat{j}+2\hat{k})+\lambda(\hat{i}-\hat{j}+2\hat{k})$, $L_2:\vec{r}=(\hat{j}-\hat{k})+\mu(3\hat{i}+\hat{j}+p\hat{k})$ and $L_3:\vec{r}=\delta(l\hat{i}+m\hat{j}+n\hat{k})$ be three lines such that $L_1$ is perpendicular to $L_2$ and $L_3$ is perpendicular to both $L_1$ and $L_2$. Then the point which lies on $L_3$ is
Step-by-Step Solution
Key Concept: $L_1\perp L_2$: $(1,-1,2)\cdot(3,1,p)=0\Rightarrow3-1+2p=0\Rightarrow p=-1$. Direction of $L_3=(l,m,n)\perp(1,-1,2)$ and $\perp(3,1,-1)$: $(l,m,n)=((-1)(-1)-(2)(1),(2)(3)-(1)(-1),(1)(1)-(-1)(3))=(1-2,6+1,1+3)=(-1,7,4)$.
$(l,m,n)=(-1,7,4)$. Point $(-1,7,4)$ lies on $L_3$.
Correct Answer: 1