The ratio in which the plane $\bar{r} \cdot (\hat{i}-2 \hat{j}+3 \hat{k})=17$ divides the line joining the points $-2 \hat{i}+4 \hat{j}+7 \hat{k}$ and $3 \hat{i}-5 \hat{j}+8 \hat{k}$ is:

If the lines $L_1 : \frac{x - 1}{-3} = \frac{y - 2}{2k} = \frac{z - 3}{2}$ and $L_2 : \frac{x - 1}{3k} = \frac{y - 5}{1} = \frac{z - 6}{-5}$ are perpendicular to each other, then the equation of a plane containing the line $L_1$ and parallel to the line $L_2$ for the value of $k$ that satisfies this condition is

Show that the lines $\frac{x-a+d}{\alpha-\delta}=\frac{y-a}{\alpha}=\frac{z-a-d}{\alpha+\delta}$ and $\frac{x-b+c}{\beta-\gamma}=\frac{y-b}{\beta}=\frac{z-b-c}{\beta+\gamma}$ are coplanar.

The distance of the origin from the plane $r \cdot(3 \hat{i}+4 \hat{j}-12 \hat{k})=7$ measured parallel to the line $r=(\hat{i}+2 \hat{j}+3 \hat{k})+t(6 \hat{i}+2 \hat{j}+3 \hat{k})$ is

The equation of the plane passing through the point $(2,-1,-3)$ and parallel to the lines $\frac{x-1}{3}=\frac{y+2}{2}=\frac{z}{-4}$ and $\frac{x}{2}=\frac{y-1}{-3}=\frac{z-2}{2}$ is

The distance of the point $(3, 8, 2)$ from the line $\frac{x - 1}{2} = \frac{y - 3}{4} = \frac{z - 2}{3}$ measured parallel to the plane $3x + 2y - 2z = 0$ is