(a) If [tex](x,0,0)[/tex] lies on both planes, then
[tex]y=z=0 \implies 3x = 3 \implies \boxed{x=1}[/tex]
and at the same time
[tex]2x = 2 \implies x=1[/tex]
(b) A plane with normal vector [tex]\vec n[/tex] containing the point [tex](a,b,c)[/tex] can be written in the form
[tex]\vec n \cdot \langle x - a, y - b, z - c \rangle = 0[/tex]
Expanding the left side, we see that the components of [tex]\vec n[/tex] correspond to the coefficients of [tex]x,y,z[/tex]. So the normal vector to [tex]3x-6y-2z[/tex] is [tex]\vec n_1 = \boxed{\langle3,-6,-2\rangle}[/tex].
(c) Similarly, the normal to [tex]2x+y-2z=2[/tex] is [tex]\vec n_2 = \boxed{\langle2,1,-2\rangle}[/tex].
(d) The cross product of any two vectors [tex]\vec x[/tex] and [tex]\vec y[/tex] is perpendicular to both of the vectors. So we have
[tex]\vec n_1 \times \vec n_2 = \boxed{\langle 14, 2, 15\rangle}[/tex]
(e) Solve the two plane equations for [tex]z[/tex].
[tex]3x - 6y - 2z = 3 \implies 2z = 3x - 6y - 3[/tex]
[tex]2x + y - 2z = 2 \implies 2z = 2x + y - 2[/tex]
By substitution,
[tex]3x - 6y - 3 = 2x + y - 2 \implies x = 7y + 1[/tex]
Let [tex]y=t\in\Bbb R[/tex]. Then [tex]x=7t+1[/tex] and
[tex]2z = 3(7t+1) - 6t - 3 \implies 2z = 15t \implies z = \dfrac{15}2t[/tex]
Then the intersection can be parameterized by equations
[tex]\begin{cases} x(t) = 7t + 1 \\\\ y(t) = t \\\\ z(t) = \dfrac{15}2 t \end{cases}[/tex]
for [tex]t\in\Bbb R[/tex].
We can also set [tex]x=t[/tex] or [tex]z=t[/tex] first, then solve for the other variables in terms of the parameter [tex]t[/tex], so this is by no means a unique parameterization.
