Hint
If the plane is parallel to the lines given by $\vec x = \vec p + \lambda \color{blue}{\vec v}$ and $\vec x = \vec q + \lambda\color{blue}{ \vec w}$, then $\vec n = \color{blue}{\vec v} \times \color{blue}{\vec w}$ is a normal vector of the plane. A Cartesian equation of the plane with normal vector $\vec n = (a,b,c)$ and containing a point $(x_0,y_0,z_0)$ is given by:
$$a\left( x-x_0 \right)+b\left( y-y_0 \right)+c\left( z-z_0 \right)=0$$
(A),(B),(C): Looks right.
The problem is , the direction vector I found for $L$ is $(1,-1,-1)$, which is equal to the one for $\Pi$! Where did I made mistakes? Can anyone help me ??
The equation for $\Pi$ is, as you found, $x-y-z=2$, or equivalently $(x,y,z)\cdot (1,-1,-1)=2$. That vector $(1,-1,-1)$? It's perpendicular to the plane, not any of the directions in it. You haven't made any calculation mistakes here; it's just a question of how to interpret the plane's equation.
(E): The hydrogen atoms in a methane molecule form a regular tetrahedron, centered at the carbon atom. If we know the positions of three of the points $A,B,C$, the fourth point $D$ lies on the line through the center of triangle $ABC$ perpendicular to the plane those points are on, and also satisfies $AD=AB$. There are of course two points on the line that satisfy this, and both of them work in that they're possible fourth vertices of the tetrahedron.
So, then, there are two things we need to verify about that point $(-1,4,5)$: the distance equality, and that it's on the line we found in part (D).
(F): We need to find the center of that regular tetrahedron. Building on what we already know, it will be a point equidistant from all four vertices. We found planes in part (C) that were the locus of points equidistant from $A$ and $B$ and equidistant from $A$ and $C$ respectively. We found a line in part (D) that was the intersection of those planes - the locus of all points equidistant from $A,B,C$. To find the point equidistant from all four points $A,B,C,D$, we should intersect this line with one more plane, perpendicular to segment $AD$ through its midpoint. (Or, since the statement gave you a set of coordinates, we could just test those for equal distances.)
Best Answer
for part d), i'm not 100% sure but couldn't we say that it will either be the normal vector n, or the one in the opposite direction -n? and then use the formula for a distance (d) from a point P (the origin) to a plane containing Q (found from part a) with normal vector (n) as d = |n.PQ|/|n|, but then take away the absolute values from the top part of the fraction to give us a "direction" to the distance between the plane and the origin and thus the unit vector will be the one with the same sign as that direction? like i said this could be total rubbish, but just a thought! :)