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:

Find the equation of the plane passing through the intersection of the planes $x + 2y + 3z - 4 = 0$ and $2x + y - z + 5 = 0$ and perpendicular to the plane $5x + 3y + 6z + 8 = 0$.

If ${L_1}$ is the line of intersection of the planes $2x - 2y + 3z - 2 = 0$ and $x - y + z + 1 = 0$,and ${L_2}$ is the line of intersection of the planes $x + 2y - z - 3 = 0$ and $3x - y + 2z - 1 = 0$,then the distance of the origin from the plane containing the lines ${L_1}$ and ${L_2}$ is:

If the line of intersection of the planes $2x + 3y + z = 1$ and $x + 3y + 2z = 2$ makes an angle $\alpha$ with the positive $x$-axis,then $\cos \alpha = $

The position vector of the point of intersection of the line $\bar{r}=(2 \hat{i}+\hat{j}-4 \hat{k})+\lambda(\hat{i}-2 \hat{j}+2 \hat{k})$ and the $XOY$-plane is:

The ratio in which the line joining $(2, -4, 3)$ and $(-4, 5, -6)$ is divided by the plane $3x + 2y + z - 4 = 0$ is