The value of $k$,such that the line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}$ lies on the plane $2x-4y+z=7$,is

The vector equation of any plane passing through the line of intersection of the planes $\vec{r} \cdot \vec{m}_1=q_1$ and $\vec{r} \cdot \vec{m}_2=q_2$ is given by $\vec{r} \cdot (\vec{m}_1+\lambda \vec{m}_2)=q_1+\lambda q_2$ for $\lambda \in R$. Find the vector equation of the plane passing through the point $2 \hat{i}-3 \hat{j}+\hat{k}$ and the line of intersection of the planes $\vec{r} \cdot (\hat{i}-2 \hat{j}+3 \hat{k})=5$ and $\vec{r} \cdot (3 \hat{i}+\hat{j}-2 \hat{k})=7$.

The equation of the plane passing through the point $\hat{i}+2 \hat{j}-\hat{k}$ and perpendicular to the line of intersection of the planes $r \cdot(3 \hat{i}-\hat{j}+\hat{k})=1$ and $r \cdot(\hat{i}+4 \hat{j}-2 \hat{k})=2$ is:

The equation of the plane passing through the lines $\frac{x - 4}{1} = \frac{y - 3}{1} = \frac{z - 2}{2}$ and $\frac{x - 3}{1} = \frac{y - 2}{-4} = \frac{z}{5}$ is

The image of the line $\frac{x - 1}{3} = \frac{y - 3}{1} = \frac{z - 4}{-5}$ in the plane $2x - y + z + 3 = 0$ is the line:

If the distance between the plane $Ax-2y+z=d$ and the plane containing the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ is $\sqrt{6}$ units,then $|d|$ is