If the line $\frac{x - 4}{1} = \frac{y - 2}{1} = \frac{z - k}{2}$ lies on the plane $2x - 4y + z = 7$,then $k = . . . . $

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 origin and containing the line $\frac{x - 1}{5} = \frac{y - 2}{4} = \frac{z - 3}{5}$ is:

The sine of the angle between the straight line $\frac{x-2}{2}=\frac{y-3}{4}=\frac{4-z}{2}$ and the plane $2x-2y+z=5$ is

$A$ plane $\pi$ passing through the points $2 \hat{i}-3 \hat{j}$ and $3 \hat{i}+4 \hat{k}$ is parallel to the vector $2 \hat{i}+3 \hat{j}-4 \hat{k}$. If a line joining the points $\hat{i}+2 \hat{j}$ and $\hat{j}-2 \hat{k}$ intersects the plane $\pi$ at the point $a \hat{i}+b \hat{j}+c \hat{k}$, then $a+b+2c=$

The line $\frac{x - 3}{2} = \frac{y - 4}{3} = \frac{z - 5}{4}$ lies in the plane $4x + 4y - kz - d = 0$. The values of $k$ and $d$ are