Under what condition is the straight line $\frac{x - x_0}{l} = \frac{y - y_0}{m} = \frac{z - z_0}{n}$ parallel to the $xy$-plane?

In $\mathbb{R}^3$,let $L$ be a straight line passing through the origin. Suppose that all the points on $L$ are at a constant distance from the two planes $P_1: x+2y-z+1=0$ and $P_2: 2x-y+z-1=0$. Let $M$ be the locus of the feet of the perpendiculars drawn from the points on $L$ to the plane $P_1$. Which of the following points lie$(s)$ on $M$?
$(A) \left(0, -\frac{5}{6}, -\frac{2}{3}\right)$
$(B) \left(-\frac{1}{6}, -\frac{1}{3}, \frac{1}{6}\right)$
$(C) \left(-\frac{5}{6}, 0, \frac{1}{6}\right)$
$(D) \left(-\frac{1}{3}, 0, \frac{2}{3}\right)$

If a plane $P$ passes through the points $(1,0,0)$ and $(0,1,0)$ and makes an angle $\frac{\pi}{4}$ with the plane $x+y=3$,then the direction ratios of a normal to that plane $P$ are

Let $L$ be the line of intersection of the planes $2x + 3y + z = 1$ and $x + 3y + 2z = 2$. If $L$ makes an angle $\alpha$ with the positive direction of the $x$-axis,then $\cos \alpha$ is:

The position vector of the point where the line $r = i - j + k + t(i + j - k)$ meets the plane $r \cdot (i + j + k) = 5$ is

If the shortest distance between the lines $\frac{x - 1}{\alpha} = \frac{y + 1}{-1} = \frac{z}{1}, (\alpha \ne -1)$ and $x + y + z + 1 = 0 = 2x - y + z + 3$ is $\frac{1}{\sqrt{3}}$,then a value of $\alpha$ is