The line passing through the points $(1, 1, -1)$ and $(3, -1, 0)$ makes an angle of $\operatorname{Tan}^{-1}\left(\frac{1}{\sqrt{8}}\right)$ with the plane $\sqrt{\lambda} x + 3y + 6z = 17$. Then $\lambda =$

The equation of the plane passing through the point $\hat{i}+2 \hat{j}-\hat{k}$ and perpendicular to the line of intersection of the planes $r \cdot(3 \hat{i}-\hat{j}+\hat{k})=1$ and $r \cdot(\hat{i}+4 \hat{j}-2 \hat{k})=2$ is:

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.

The symmetric equation of the line formed by the intersection of the planes $3x + 2y + z - 5 = 0$ and $x + y - 2z - 3 = 0$ is:

From a point $P(\lambda, \lambda, \lambda)$,perpendiculars $PQ$ and $PR$ are drawn respectively on the lines $y=x, z=1$ and $y=-x, z=-1$. If $P$ is such that $\angle QPR$ is a right angle,then the possible value$(s)$ of $\lambda$ is(are)

Three lines $L_1: \overrightarrow{r} = \lambda \hat{i}, \lambda \in R$,$L_2: \overrightarrow{r} = \hat{k} + \mu \hat{j}, \mu \in R$,and $L_3: \overrightarrow{r} = \hat{i} + \hat{j} + v\hat{k}, v \in R$ are given. For which point$(s)$ $Q$ on $L_2$ can we find a point $P$ on $L_1$ and a point $R$ on $L_3$ such that $P, Q,$ and $R$ are collinear?