If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-\lambda}{2}=\frac{z}{1}$ intersect each other,then $\lambda = \ldots$

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:

If the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - c}{2} = \frac{z}{1}$ intersect, then the radius of the circle $x^2 + y^2 - 4x + 10y + c = 0$ is

The lines $\frac{x - 1}{-1} = \frac{y + 2}{1} = \frac{z - 3}{-2}$ and $\frac{x - 1}{1} = \frac{y + 2}{1} = \frac{z + 1}{-2}$ are

Let $A(\alpha, 4, 7)$ and $B(3, \beta, 8)$ be two points in space. If the $YZ$ plane and $ZX$ plane respectively divide the line segment joining the points $A$ and $B$ in the ratio $2:3$ and $4:5$, then the point $C$ which divides $\overline{AB}$ in the ratio $\alpha: \beta$ externally is

$A$ line $L$ passes through points $A(1, 3, 2)$ and $B(2, 2, 1)$. If the mirror image of point $P(1, 1, -1)$ in the line $L$ is $(x, y, z)$,then $x+y+z=$