Equation of a Line in 3D

The distance of the point $(1,2,3)$ from the plane $x+y+z=2$ measured parallel to the line $\frac{x+1}{-1}=\frac{y}{-2}=\frac{z-3}{1}$ is

If the lines $\frac{\text{x}-1}{2}=\frac{\text{y}+1}{3}=\frac{\text{z}-1}{4}$ and $\frac{\text{x}-3}{1}=\frac{\text{y}-\text{k}}{2}=\frac{\text{z}}{1}$ intersect, then the value of k is

The equation of line equally inclined to co-ordinate axes and passing through $\left(-3,2,-5\right)$ is

The vector equation of line passing through the point $\left(5,4,3\right)$ and having direction ratios $-3,4,2$ is

If the points $A(-1,3,2), B(-4,2,-2)$ and $\mathrm{C}(5,5,\lambda )$ are collinear, find the value of $\lambda$.

Find the vector equation of a line passing through the point $\hat{i}+2\hat{j}+3\hat{k}$, and perpendicular to the vectors $\hat{i}+\hat{j}+\hat{k}$ and $2\hat{i}-\hat{j}+\hat{k}$.

If the Cartesian equation of a line is $\frac{x-6}{2}=\frac{y+4}{7}=\frac{z-5}{3}$, then the vector equation of the line is

Equation of line equally inclined to coordinate axes and passes through $\left(-5,1,-2\right)$ is

If the points $A\left(5,5,\lambda \right),B\left(-1,3,2\right)$ and $C\left(-4,2,-2\right)$ are collinear, then the value of $\lambda$ is

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

The vector equation of the line through the points $(3, 4, -7)$ and $(1, -1, 6)$ is

If the points $A\left(5,5,\lambda \right), B\left(-1,3,2\right)$ and $C\left(-4,2,-2\right)$ are collinear, find the value of $\lambda$.

Find the vector equation of the line passing through the points $(3,4,-7)$ and $(6,-1,1)$

Equation of the line passing through the point $(3,1,2)$ and perpendicular to the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\frac{x}{3}=\frac{y}{2}=\frac{z}{5}$ is

If the point $A(\lambda ,5,-2)$ lies on the line $\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}$, then the value of $\lambda$ is