Distance between the point (3, 4, 5) and the point where the line $\frac{x-3}{1} = \frac{y-4}{2} = \frac{z-5}{2}$ meets the plane $x + y + z = 17$ is :

Match List-I with List-II

Choose the correct answer from the options given below:

If the shortest distance between the lines $l_1$ and $l_2$ given by $\vec{r} = a\hat{i} + 2\hat{j} - \hat{k} + \lambda(2\hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = \hat{i} - \hat{j} + \hat{k} + \mu(2\hat{i} - \hat{j} + \hat{k})$ is $\sqrt{\frac{35}{6}}$ units, the values of 'a' can be :

The shortest distance between lines $\frac{-x-3}{4} = \frac{y-6}{3} = \frac{z}{2}$ and $\frac{-x-2}{4} = \frac{y}{1} = \frac{z-7}{1}$ is:

Equation of y-axis in space, in vector form is :

The coordinates of the image of the point P (5, 4, 2) in the line $\vec{r} = (-\hat{i} + 3\hat{j} + \hat{k}) + \mu(2\hat{i} + 3\hat{j} - \hat{k})$, where $\lambda$ is a parameter, is

The distance of the plane $\vec{r} \cdot \left(\frac{2}{7}\hat{i} + \frac{3}{7}\hat{j} - \frac{6}{7}\hat{k}\right) = 2$ from the origin is