The line segment joining the points $A(2, 3, 4)$ and $B(-3, 5, -4)$ intersects the $yz$-plane at the point:

Given the lines $\vec{r} = (3+t)\hat{i} + (1-t)\hat{j} + (-2-2t)\hat{k}$,$t \in R$ and $x = 4+k, y = -k, z = -4-2k$,$k \in R$. What is the relationship between these two lines?

The equation of the line of the shortest distance between the lines $\frac{x}{1} = \frac{y}{-1} = \frac{z}{1}$ and $\frac{x - 1}{0} = \frac{y + 1}{-2} = \frac{z}{1}$ is

If a point $R(4, y, z)$ lies on the line segment joining the points $P(2, -3, 4)$ and $Q(8, 0, 10)$,then the distance of $R$ from the origin is

If $P$ is a point lying on the line passing through the point $A(\hat{i}-\hat{j}+3 \hat{k})$ and parallel to the vector $2 \hat{i}+\hat{j}-2 \hat{k}$ such that $|AP|=18$,then a position vector of $P$ is

Let the line passing through the points $(-1,2,1)$ and parallel to the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$ intersect the line $\frac{x+2}{3}=\frac{y-3}{2}=\frac{z-4}{1}$ at the point $P$. Then the distance of $P$ from the point $Q(4,-5,1)$ is: