51.

The point of intersection of the straight lines $\mathrm{r} = \left(3\hat{\mathrm{i}} - 4\hat{\mathrm{j}} + 5\hat{\mathrm{k}}\right) + \mathrm{\lambda }\left(- \hat{\mathrm{i}} - 2\hat{\mathrm{j}} + 2\hat{\mathrm{k}}\right)$ and $\frac{3 - \mathrm{x}}{- 1} = \frac{\mathrm{y} + 4}{2} = \frac{\mathrm{z} - 5}{7}$ is

• (- 3, - 4, - 5)

• (- 3, 4, 5)

• (- 3, 4, - 5)

• (3, - 4, 5)

The vector equation of the straight line $\frac{\mathrm{x} - 2}{- 1} = \frac{\mathrm{y}}{- 3} = \frac{1 - \mathrm{z}}{2}$ is

• $\mathrm{r} = \left(2\hat{\mathrm{i}} + \hat{\mathrm{k}}\right) + \mathrm{t}\left(\hat{\mathrm{i}} + 3\hat{\mathrm{j}} + 2\hat{\mathrm{k}}\right)$

• $\mathrm{r} = \left(2\hat{\mathrm{i}} - \hat{\mathrm{k}}\right) + \mathrm{t}\left(\hat{\mathrm{i}} - 3\hat{\mathrm{j}} - 2\hat{\mathrm{k}}\right)$

• $\mathrm{r} = \left(2\hat{\mathrm{i}} + \hat{\mathrm{k}}\right) + \mathrm{t}\left(\hat{\mathrm{i}} - 3\hat{\mathrm{j}} + 2\hat{\mathrm{k}}\right)$

• $\mathrm{r} = \left(2\hat{\mathrm{i}} + \hat{\mathrm{k}}\right) + \mathrm{t}\left(\hat{\mathrm{i}} - 3\hat{\mathrm{j}} - 2\hat{\mathrm{k}}\right)$

The straight line $\mathrm{r} = (\hat{\mathrm{i}} + \hat{\mathrm{j}} + 2\hat{\mathrm{k}}) + \mathrm{t}\left(2\hat{\mathrm{i}} + 5\hat{\mathrm{j}} + 3\hat{\mathrm{k}}\right)$ is parallel to the plane $\mathrm{r} . (2\hat{\mathrm{i}} + \hat{\mathrm{j}} - 3\hat{\mathrm{k}}) = 5$. Then, the distance between the straight line and the plane is

• $\frac{9}{\sqrt{14}}$

• $\frac{8}{\sqrt{14}}$

• $\frac{7}{\sqrt{14}}$

• $\frac{6}{\sqrt{14}}$

Two fair dice are rolled. Then, the probability of getting a composite number as the sum of face values is equal to

• $\frac{7}{12}$

• $\frac{5}{12}$

• $\frac{1}{12}$

• $\frac{3}{4}$