Equation of a Line in 3D

If the lines ($\frac{x+1}{1}=\frac{y+2}{\lambda }=\frac{z-1}{-1}$) and ($\frac{x-1}{-\lambda }=\frac{y+1}{-2}=\frac{z+1}{1}$) are mutually perpendicular, then what is the value of $\lambda$?

The equation of the straight line passing through $\left(1,\mathrm{ }2,\mathrm{ }3\right)$ and perpendicular to the plane $x+2y-5z+9=0$ is

The equation of the straight line passing through the points $(4, -5, -2)$ and $(-1, 5$, $3)$ is

The equation of the line passing through the point $\left(3, 0, -4\right)$ and perpendicular to the plane $2x-3y+5z-7=0$ is

The equation of the perpendicular from the point $\left(\alpha , \beta , \gamma \right)$ to the plane $ax+by+cz+d=0$ is

The two lines $x=ay+b, z=cy+d$ and $x=a^{\prime }y+b^{\prime }, z=c^{\prime }y+d$ are perpendicular to each other, if

The vector equation of the line passing through the points $\left(3,2,1\right)$ and $\left(-2,1,3\right)$is

Cartesian form of the equation of line $\overrightarrow{\mathbf{r}}=3\widehat{\mathbf{i}}-5\widehat{\mathbf{j}}+7\widehat{\mathbf{k}}+\lambda (2\widehat{\mathbf{i}}+\widehat{\mathbf{j}}-3\widehat{\mathbf{k}})$ is