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

Let L be the foot of the perpendicular drawn from the point $P(5,3 k-7,-4)$ to the YZ - plane. If the distance of point L from the origin is $\sqrt{41}$ units, then the possible value of ' $\boldsymbol{k}$ ' is:

Let $\mathbf{P}$ be a point on the line $L_1: \frac{x-2}{2}=y+1=\frac{z-1}{2}$ such that its distance from the point $A(2,-1,1)$ is 6 units.

Given that $\boldsymbol{x}$-coordinate of $\mathbf{P}$ is greater than $\mathbf{2}$,

Find the coordinates of point Q on the line $L_2: x-1=\frac{y-2}{2}=\frac{z-2}{2}$ such that $\mathbf{Q}$ is the closest point to $\mathbf{P}$.