The image of the point with position vector $(\hat{i}+3 \hat{j}+4 \hat{k})$ in the plane $r \cdot(2 \hat{i}-\hat{j}+\hat{k})+3=0$ is

$\vec{r} \cdot(\hat{i}-\hat{j}+\hat{k})=5$ and $\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})=3$ are two planes. $A$ plane $\pi$ passing through the line of intersection of these two planes, passes through the point $(0,1,2)$. If the equation of $\pi$ is $\vec{r} \cdot(a \hat{i}+b \hat{j}+c \hat{k})=m$, then $\frac{b c}{a^2}=$

The equation of the plane passing through the line of intersection of the planes $x+y+z=1$ and $3x+4y+5z=2$ and perpendicular to the $XY$-plane is

The lines $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-K}$ and $\frac{x-1}{K}=\frac{y-4}{2}=\frac{z-5}{1}$ are coplanar if

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 $\vec{r} \cdot (3 \hat{i}+\hat{j}-2 \hat{k})=7$.

If the lines $x = 1 + s, y = -3 - \lambda s, z = 1 + \lambda s, s \in R$ and $x = \frac{t}{2}, y = 1 + t, z = 2 - t, t \in R$ are coplanar,then $\lambda = $