If $\mathbf{r}=(2-\lambda+\mu) \hat{\mathbf{i}}+(1-\mu) \hat{\mathbf{j}}+(2-3 \lambda+2 \mu) \hat{\mathbf{k}}$ is the vector equation of a plane, then the equivalent cartesian equation of the plane is

Let $\pi_1$ be a plane passing through the point $\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and perpendicular to the vector $-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$. Let the line $L$ passing through the points $3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $-\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ be a normal to the plane $\pi_2$. If the angle between the planes $\pi_1$ and $\pi_2$ is $\theta$, then $\cos \theta=$

Let $A=(1,2,0), B=(2,0,-1), C=(0,-2,3)$ and $D=(-1,2,-3)$ be four points in the space. Let $G_1$ be the centroid of $\triangle A B C$ and $G_2$ be the centroid of tetrahedron $A B C D$. If $P$ divides, $G_1 G_2$ in the ratio $4: 3$ internally, then $P=$

If the d.r.'s of two lines are connected by the relations $a-b+c=0, a^2-b^2+2 c^2=0$ and $\theta$ is the angle between these lines, then $\cos \theta=$