The point $P$ lies on the line $AB$,where $A \equiv (2, 4, 5)$ and $B \equiv (1, 2, 3)$. If the $z$-coordinate of point $P$ is $3$,then its $y$-coordinate is:

Let $L_1: \frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}$ and $L_2: \frac{x-2}{2}=\frac{y}{0}=\frac{z+4}{\alpha}, \alpha \in R$,be two lines,which intersect at the point $B$. If $P$ is the foot of perpendicular from the point $A(1,1,-1)$ on $L_2$,then the value of $26 \alpha(PB)^2$ is . . . . . . .

The perpendicular distance from the point $P(-1, 1, 0)$ to the line joining the points $A(0, 2, 4)$ and $B(3, 0, 1)$ is

The angle between a line with direction ratios $2 : 2 : 1$ and a line joining $(3, 1, 4)$ to $(7, 2, 12)$ is

Find the equation of the line in vector and in Cartesian form that passes through the point with position vector $2\hat{i}-\hat{j}+4\hat{k}$ and is in the direction $\hat{i}+2\hat{j}-\hat{k}$.

If for some $m \in \mathbb{R}$ the lines $L_1 : \frac{x + 1}{m} = \frac{y - m}{-1} = \frac{z - 1}{1}$ and $L_2 : \frac{x + 2}{-4} = \frac{y + 1}{9} = \frac{z + 1}{1}$ are coplanar, then line $L_1$ passes through the point