The equation of the perpendicular drawn from the point $A(6,1,3)$ to the line $\frac{x-1}{2}=\frac{2-y}{-1}=\frac{z-3}{2}$ is $\frac{x-6}{\boldsymbol{a}}=\frac{y-1}{\boldsymbol{b}}=\frac{z-3}{\boldsymbol{c}}$. If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are the possible integers such that $\boldsymbol{a}<0$, then the value of $\boldsymbol{a}-\boldsymbol{b}+\mathbf{5} \boldsymbol{c}$ is:

$$ \begin{aligned} &\text { Consider two skew lines in 3D space. }\\ &M_1: \frac{x-1}{1}=\frac{2-y}{1}=\frac{z-5}{1} \text { and } M_2: \frac{x+3}{1}=\frac{y-7}{2}=\frac{z+4}{1} \end{aligned} $$

Let $L_1$ be the line of shortest distance (common perpendicular) between $M_1$ and $M_2$

If $L_2$ is a line parallel to the vector $\vec{b}=\hat{\jmath}+\hat{k}$,

Then the acute angle $\boldsymbol{\theta}$ between the lines $L_1$ and $L_2$ is:

Consider the lines $L_1$ and $L_2$ given by the following vector equations:

$$ L_1: \vec{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(3 \hat{i}+\boldsymbol{t} \hat{j}) \quad L_2: \vec{r}=(4 \hat{i}+\boldsymbol{a} \hat{j}-\hat{k})+\mu(2 \hat{i}+3 \hat{k}) $$

If $\boldsymbol{a}=-2$ and the lines intersect, then the value of ' $\mathbf{t}$ ' is:

The angle between the two lines whose direction cosines satisfy the relations $\boldsymbol{l}+\boldsymbol{m}+\boldsymbol{n}=\mathbf{0}$ and $\boldsymbol{l}^{\mathbf{2}}=\boldsymbol{m}^{\mathbf{2}}+\boldsymbol{n}^{\mathbf{2}}$ is