If $(a, b, c)$ are the direction ratios of a line joining the points $(4, 3, -5)$ and $(-2, 1, -8)$,then the point $P(a, 3b, 2c)$ lies on the plane:

If $A(1,2,3), B(3,7,-2), C(6,7,7)$ and $D(-1,0,-1)$ are points in a plane, then the vector equation of the line passing through the centroids of $\triangle ABD$ and $\triangle ACD$ is

The equation of the line passing through the point $(1, 2, -4)$ and perpendicular to the two lines $\frac{x - 8}{3} = \frac{y + 19}{-16} = \frac{z - 10}{7}$ and $\frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{-5}$ is:

The angle between the lines $\vec{r}=(2 \hat{i}+\hat{j}-3 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})$ and $\frac{x-1}{1}=\frac{y+2}{3}=\frac{z-3}{2}$ is

The lines $\frac{x-3}{1}=\frac{y-2}{1}=\frac{z-5}{-k}$ and $\frac{x-4}{k}=\frac{y-3}{1}=\frac{z-3}{2}$ are coplanar,hence $k=$

The graph of the equation $y^2 + z^2 = 0$ in three-dimensional space is