If the point of intersection of the lines $r = \hat{i} - 6\hat{j} + (p \sec \alpha) \hat{k} + t(\hat{i} + 2\hat{j} + \hat{k})$ and $r = 4\hat{j} + \hat{k} + \lambda(2\hat{i} + (p \tan \alpha) \hat{j} + 2\hat{k})$ is $8\hat{i} + 8\hat{j} + 9\hat{k}$, (where $0 < \alpha < \frac{\pi}{2}$), then $p =$

The angle made by the vector $2\hat{i}-\hat{j}+\hat{k}$ with the plane represented by $\vec{r} \cdot(\hat{i}+\hat{j}+2\hat{k})=7$ is (in $^{\circ}$)

The plane $lx + my = 0$ is rotated by an angle $\alpha$ about its line of intersection with the plane $z = 0$. Find the equation of the plane in its new position.

Let $\pi_1$ be the plane determined by the vectors $\bar{i}+\bar{j}$ and $\bar{i}+\bar{k}$, and $\pi_2$ be the plane determined by the vectors $\bar{j}-\bar{k}$ and $\bar{k}-\bar{i}$. Let $\bar{a}$ be a non-zero vector parallel to the line of intersection of the planes $\pi_1$ and $\pi_2$. If $\bar{b}=\bar{i}+\bar{j}-\bar{k}$, then the angle between the vectors $\bar{a}$ and $\bar{b}$ is:

The angle $\theta$ between the line $x = \frac{y-1}{2} = \frac{z-3}{\lambda}$ and the plane $x + 2y + 3z = 6$ is given by $\cos^{-1} \sqrt{\frac{5}{14}}$. Find the value of $\lambda$.

Let $P(2,1,5)$ be a point in space and $Q$ be 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 $3x-y+4z=1$ is