So in basic geometry, when lines are parallel they never intersect i.e have solutions.

Considering, there is no solution for x?

It means that all the planes are parallel

So the answer would be 3/C?

Here we have three planes:

\(\displaystyle \begin{align*}\text{ i. }x+4y+6z&=18 \\\text{ ii. }x-2y+z&=-6\\\text{iii. }2x+4y+17z&=-6 \end{align*}\)

Because no two of those planes are parallel then each pair have a line of intersection.

Planes plane i. has normal \(\displaystyle N_1=<1.4.6>\) likewise \(\displaystyle N_2=<1,-2,1>\text{ & }N_3=<2,14,17>\).

Then \(\displaystyle N_1\times N_2=<16,-5,-6\) is the direction vector for the the intersection of planes \(\displaystyle \text{ i. & ii. }\).

Likewise, \(\displaystyle N_1\times N_3=<-16,5,6>\) is the direction vector for the the intersection of planes \(\displaystyle \text{ i. & iii. }\).

Because those two lines are parallel and we know that there is no point common to all three planes what is your answer.?