If points $P(4, 5, x)$,$Q(3, y, 4)$,and $R(5, 8, 0)$ are collinear,then the value of $x+y$ is

Let the line $L$ intersect the lines $x-2=-y=z-1$ and $2(x+1)=2(y-1)=z+1$,and be parallel to the line $\frac{x-2}{3}=\frac{y-1}{1}=\frac{z-2}{2}$. Then which of the following points lies on $L$?

Line $L_1$ passes through the points $\hat{i}+\hat{j}$ and $\hat{k}-\hat{i}$. Line $L_2$ passes through the point $\hat{j}+2\hat{k}$ and is parallel to the vector $\hat{i}+\hat{j}+\hat{k}$. If $x\hat{i}+y\hat{j}+z\hat{k}$ is the point of intersection of the lines $L_1$ and $L_2$,then $(y-x)=$

The line,that is coplanar to the line $\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}$,is

The equation of the line passing through the point $(0,1,2)$ and perpendicular to the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2}$ is

Statement-$1$: The distance between the two parallel lines $\frac{x}{2} = \frac{y}{-1} = \frac{z}{2}$ and $\frac{x-1}{4} = \frac{y-1}{-2} = \frac{z-1}{4}$ is $\sqrt{2}$.
Statement-$2$: The distance between two parallel lines is equal to the perpendicular distance from any point on one line to the other line.