If the angle $\theta$ between the line $\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}$ and the plane $2x-y+\sqrt{\lambda}z+4=0$ is such that $\sin \theta=\frac{1}{3}$,then the value of $\lambda=$

Let the foot of the perpendicular from a point $P(1,2,-1)$ to the straight line $L: \frac{x}{1}=\frac{y}{0}=\frac{z}{-1}$ be $N$. Let a line be drawn from $P$ parallel to the plane $x+y+2z=0$ which meets $L$ at point $Q$. If $\alpha$ is the acute angle between the lines $PN$ and $PQ$,then $\cos \alpha$ is equal to $.....$

The square of the distance of the point of intersection of the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+1}{6}$ and the plane $2x-y+z=6$ from the point $(-1,-1,2)$ is .... .

The equation of the plane passing through the line of intersection of the planes $x + y + z = 1$ and $2x + 3y - z + 4 = 0$ and parallel to the $x$-axis is:

Let $P$ be the plane passing through the intersection of the planes $\overrightarrow{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) = 5$ and $\overrightarrow{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 3$,and the point $(2, 1, -2)$. Let the position vectors of the points $X$ and $Y$ be $\hat{i} - 2\hat{j} + 4\hat{k}$ and $5\hat{i} - \hat{j} + 2\hat{k}$ respectively. Then the points:

The equation of the plane containing the line $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$ and perpendicular to the plane containing the lines $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$ is