The equation of the plane containing the line $\frac{x+1}{2}=\frac{y+2}{1}=\frac{z-2}{3}$ and the point $(1,-1,3)$ is

The line L is passing through $(1,2,3)$. The distance of any point on the line L from the line $\overline{\mathrm{r}}=(3 \lambda-1) \hat{\mathrm{i}}+(-2 \lambda+3) \hat{\mathrm{j}}+(4+\lambda) \hat{\mathrm{k}}$ is constant. Then the line L does not pass through the point

The distance of the plane $\overline{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}})+\lambda(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})+\mu(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$ from the origin is

If the angle between the line $x=\frac{y-1}{2}=\frac{z-3}{\lambda}$ and the plane $x+2 y+3 z=4$ is $\cos ^{-1} \sqrt{\frac{5}{14}}$, then the value of $\lambda$ is