The Cartesian equation of the plane $\vec{r}=(2 \hat{i}-3 \hat{j})+\lambda(\hat{i}+2 \hat{j}-\hat{k})+\mu(2 \hat{i}+3 \hat{j}+\hat{k})$ is

$A$ plane $\pi$ passing through the point $3 \hat{i}-4 \hat{j}+5 \hat{k}$ is parallel to the plane which passes through the point $\hat{i}+\hat{j}-\hat{k}$ and is perpendicular to the vector $\hat{i}+2 \hat{j}-3 \hat{k}$. Then the Cartesian equation of $\pi$ is

Find the angle between the planes whose vector equations are $\vec{r} \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k})=5$ and $\vec{r} \cdot(3 \hat{i}-3 \hat{j}+5 \hat{k})=3$.

The point on the plane $2x - 2y + 4z + 5 = 0$ that is nearest to $\left(1, \frac{3}{2}, 2\right)$ is

Find the equation of the plane passing through the intersection of the planes $3x - y + 2z - 4 = 0$ and $x + y + z - 2 = 0$ and the point $(2, 2, 1)$.

Find the equation of the plane passing through the intersection of the planes $\vec{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) = 5$ and $\vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 3$,and passing through the point $(2, 1, -2)$.