Cannot figure out?

The direction vector of the line is given by coefficients of t in line equation: D = (-1,-5,2)
The directions of the *normals* to the planes are given by coefficients of x, y and z in plane equations:

(a) na = (-14,4,3)
(b) nb = (5,2,5)
(c) nc = (-1.5,-7.5,3)
(d) nd = (-21,7,7)

If the dot product of D and na to nd =0 then D (and hence the line) is parallel to the plane.
If the cross product of D and na to nd =0 then D is perpendicular to the plane.
If neither are 0 then D is neither parallel nor perpendicular to the plane.

(a)
D•na = -1*-14 + -5*4 + 2*3 = 0 ---> parallel
(b)
D•nb = -1*5 + -5*2 + 2*5 = -5
D×nb = (-5*5 - 2*2)i +..= -29i +.. ----> neither
(c)
D•nc = -1*-1.5 + -5*-7.5 + 2*3 > 0
D×nc = (-5*3 - 3*-7.5)i + (2*-1.5 - 3*-1)j + (-1*-7.5 - -5*-1.5)k = 0 ----> perpendicular
(d)
D•nd = -1*-21 + -5*7 + 2*7 = 0 ----> parallel

Hey,Boss.De plane.De plane.

This is a question on planes.

Hello I cant seem to figure out which of my answer is not correct!

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