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to a): The plane is given in normal form with $\displaystyle \vec n = (1, -2, 2)$. Re-write the equation of the plane into Hesse normal form by dividing by the value of the normal vector:
$\displaystyle |\vec n| = \sqrt{1^2+2^2+2^2} = 3$. Thus
$\displaystyle P_H:\frac13 x- \frac23 y + \frac23 z - \frac73 = 0$
Now choose a point of L1 and calculate the distance of this point to the plane. If t = 0 then Q(4, -8, 7) is a point of the line. Plug in the coordinates of Q into the equation of $\displaystyle P_H$ to calculate the distance QP:
$\displaystyle d(Q,P) = \frac13 \cdot 4- \frac23 \cdot (-8) + \frac23 \cdot 7 - \frac73 = 3$
The positive sign indicates that the point Q and the origin are on different sides of the plane P.
to b) If I understand your question correctly then you are looking for a line which is parallel to L1, is placed on the other side of P as L1 and has the same distance to P as L1. If so:
A line through A(4, -8, 7) and perpendicular to P cuts the plane in F(1, -2, 1). F is the mid point of the segment AA'. Use the midpoint formula toget the coordinates of A'(-2, 4, -5).
Thus the line L1' has the equation:
$\displaystyle \vec r = (-2,4,-5)+s\cdot (2,3,2)~\implies~\begin{array}{lr}x=&-2+2s\\y=&4+3s\\z=&-5+2s\end{array}$