Find the vector equation of the plane passing through the intersection of the planes $\vec{r} \cdot (3\hat{i} - \hat{j} + \hat{k}) = 1$ and $\vec{r} \cdot (\hat{i} + 4\hat{j} - 2\hat{k}) = 2$,and passing through the point $\hat{i} + 2\hat{j} - \hat{k}$.

If the line of intersection of the planes $ax + by = 3$ and $ax + by + cz = 0$ $(a > 0)$ makes an angle $30^{\circ}$ with the plane $y - z + 2 = 0$,then the direction cosines of the line are:

The acute angle $\theta$ between the line $\vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} + \hat{j} + \hat{k})$ and the plane $\vec{r} \cdot (2\hat{i} + p\hat{j} + \hat{k}) = 8$ is given by $\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$, where $\vec{b} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{n} = 2\hat{i} + p\hat{j} + \hat{k}$. If $\theta = \sin^{-1} \left( \frac{\sqrt{2}}{3} \right)$, then the value$(s)$ of $p$ is/are:

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

$A$ point $P$ lies on a line passing through $Q(1, -2, 3)$ and is parallel to the line $\frac{x}{1} = \frac{y}{4} = \frac{z}{5}$. If $P$ lies on the plane $2x + 3y - 4z + 22 = 0$, then the length of the segment $PQ$ is:

The equation of the plane passing through the intersection of planes $2x - y + z = 3$ and $4x - 3y + 5z = -9$ and parallel to the line $\frac{x + 1}{2} = \frac{y + 3}{4} = \frac{z - 3}{5}$ is