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

If the point $(1, \alpha, \beta)$ lies on the line of the shortest distance between the lines $\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}$ and $\frac{x+2}{-1}=\frac{y+6}{2}, \mathrm{z}=1$, then $\alpha+\beta=$