In what ratio does the plane $x - y + z = 1$ divide the line segment joining the points $(0, 0, 0)$ and $(1, -2, -5)$?

The equation of the plane containing the straight line $\frac{x}{3}=\frac{y}{2}=\frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{4}=\frac{y}{3}=\frac{z}{2}$ and $\frac{x}{2}=\frac{y}{-4}=\frac{z}{3}$ is:

The line $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ is parallel to the plane

The direction ratios of the normal to the plane passing through the points $(1, 2, -3)$, $(-1, -2, 1)$ and parallel to $\frac{x-2}{2}=\frac{y+1}{3}=\frac{z}{4}$ are:

Find the equation of the plane which contains the line of intersection of the planes $\vec{r} \cdot(\hat{i}+2 \hat{j}+3 \hat{k})-4=0$ and $\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})+5=0$ and which is perpendicular to the plane $\vec{r} \cdot(5 \hat{i}+3 \hat{j}-6 \hat{k})+8=0$.

What is the point of intersection of the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z + 2}{3}$ and the plane $2x + 3y + z = 0$?