The points on the line of intersection $\cal L$ of $\pi$ and the $xz$-plane satisfy $x + 2(0) - 2z + 3 = 0$, so $x - 2z = -3$. In particular, points $(0,0,\frac{3}{2})$, and $(1,0,2)$ are on the line, so $\cal L$ has direction vector $<1,0,\frac{1}{2}>$, or $<2,0,1>$. So the plane perpendicular to $\pi$ through $\cal L$ should have normal vector that is perpendicular to both $<1,2,-2>$ and $<2,0,1>$, and the plane passes through $(1,0,2)$, so this should be enough to finish up the problem.
So you have:
$$
L1: r=6i-3j+0k+s(3i-4j-2k)
L2: r=2i-j-4k+t(i-3j-k)
$$
Where $s$ and $t$ are the scalar parameter of your lines.
Obviously P is at the L1 line by choosing s=-1:
$$
P: 3i+j+2k
$$
PQ should be perpendicular to L1 and L2, hence the cross product between the directions will serve as new perpendicular direction:
$$
PQ: cross([3, -4, -2],[1, -3, -1])=[-2, -1, -5]
$$
Hence, you obtain the perpendicular line, starting from P. This line passes through P:
$$
L_{PQ}=P+tPQ=P: 3i+j+2k+t(-2i-1j-5k)
$$
Now we need to locate the point in $L_{PQ}$ passing through L2:
$$
L2: r=2i-j-4k+s(i-3j-k)
L_{PQ}=3i+j+2k+t(-2i-1j-5k)
$$
So:
$$
2+s=3-2t\\
-1-3s=1-t\\
-4-s=2-5t\\
$$
Solving:
$$
1+3: -2=5-7t, t=1\\
1,(1+3): s=-1\\
2,1,(1+3): 2=0
$$
Thus, a line perpendicular both to L1 and L2, passing through P do not cross with a Q in L2.
The plane passing through P and perpendicular to L1. Indeed several solutions exist, but lets take the easiest ones from the cross product of the L1 direction and the x, y and z axis.
$$
B1=cross([3, -4, -2],[1, 0, 0])=2[0, -1, 2]\\
B2=cross([3, -4, -2],[0, 1, 0])=[2, 0, 3]\\
B3=cross([3, -4, -2],[0, 0, 1])=[-4, -3, 0]
$$
All B1, B2 & B3 serve us. Hence we take B1 and B2, plus P, thus the plane is defined. Note we took B1/2:
$$
F: P+sB1+tB2= 3i+j+2k + s(0i-j+2k) + t(2i+0j+3kj)
$$
Edit: note that vectors B1 and B2 and given answer C1=[-22, -19, 5] and C2=PQ=[2 ,-1 ,5] are related through:
$$
C1=19B1-11B2\\
C2=B1+B2
$$
Thus, both C1,C2 are linearly dependent with B1,B2, hence defining the same solution:
$$
F: 3i+j+2k + s(0i-j+2k) + t(2i+0j+3kj)\\
= 3i+j+2k + (19s'+t')(0i-j+2k) + (-11s'+t')(2i+0j+3kj)\\
= 3i+j+2k + s'(-22i-19j+5k) + t'(2i-1j+5kj)
$$
Best Answer
Let L' be the desired line. We mut find a vector in the direction of L'.Since L and L' intersect, there is a plane, which we'll call $\Pi _1$, containing L and L' Similarly there is a plane,which we'll call $\Pi _2$ containing K and L'.The points (1,2,1) and (1,-3,1) are both in the plane $\Pi _1$ so we subtract their coordinates to obtain the vector [0,5,0] in the plane $\Pi _1$. The vector [1,-2,2] is also in the plane $\Pi _1$. Thus the vector $$[0,5,0] \times [1,-2,2] $$ is normal to the plane $\Pi _1$The points (1,2,1) and (2,2,0) are both in the plane $\Pi _2$ so we subtract their coordinates to obtain the vector [-1,0,1] in the plane $\Pi _2$. The vector [2,1,3] is also in the plane $\Pi _2$. Thus the vector $$[-1,0,1] \times [2,1,3] $$ is normal to the plane $\Pi _2$The line L' lies in both planes, so a vector in the direction of L', which we'll call $[a,b,c]$ must be perpendicular to the normals to both planes. Thus $$[a,b,c]=([0,5,0] \times [1,-2,2]) \times ([-1,0,1] \times [2,1,3]) $$ The equation of L' is $x=1+aT,y=2+bT,z=1+cT.$