The equation of the plane passing through the intersection of the planes $x + y + z = 6$ and $2x + 3y + 4z + 5 = 0$ and the point $(1, 1, 1)$ is:

Let $L$ be the line passing through the points $\hat{i}-9 \hat{k}$ and $7 \hat{j}+\hat{k}$ and $\pi$ be the plane passing through the point $6 \hat{i}+\hat{j}$ and perpendicular to the vector $\hat{i}+\hat{j}+\hat{k}$. If $\theta$ is the angle between $L$ and $\pi$,then $\sin \theta=$

Find the distance of the point whose position vector is $(2 \hat{i}+\hat{j}-\hat{k})$ from the plane $\vec{r} \cdot(\hat{i}-2 \hat{j}+4 \hat{k})=9$.

If $P(3, 2, 6)$ is a point in space and $Q$ is a point on the line $\vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(-3\hat{i} + \hat{j} + 5\hat{k})$,then the value of $\mu$ for which the vector $\vec{PQ}$ is parallel to the plane $x - 4y + 3z = 1$ is:

The plane passing through the points $(1, 2, 1), (2, 1, 2)$ and parallel to the line $2x = 3y, z = 1$ also passes through the point:

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 = $