Q:

When $\vec Q(\vec r) = -i\vec k \times \vec A = 0$? 

I have an equation $\vec Q(\vec r) = -i\vec k \times \vec F$, where i
  is an imaginary number and $\vec k, \vec A$ are some vectors.  I need
  to find $\vec k$ such that $\vec Q(\vec r) = 0$ so $\vec k \times \vec
F = 0$.

$\vec F(\vec r) = \vec A e^{i(\vec k \vec r-\omega t)}$ 
I have rewritten it as $(F_zk_y - F_yk_z, F_xk_z - F_zk_x, F_yk_x-F_xk_y) = (0,0,0)$, so I got three equations.  
$(F_zk_y - F_yk_z, F_xk_z - F_zk_x, F_yk_x-F_xk_y) = (A_zk_y - A_yk_z, A_xk_z - A_zk_x, A_yk_x-A_xk_y)e^{i(\vec k \vec r - \omega t)}$
It's probably trivial but I got $1 = 1$. Is it wrong or does it mean that for any $\vec k$ is $\vec Q(\vec r) = 0$?  It feels strange. 

A:

What you need to realize is that for any two non-zero  $3$-vectors $A$ and $B$, 
$A \times B = 0 \Leftrightarrow A = cB \tag{1}$
for some scalar $c$.  This follows easily from the 
definition of $A \times B$; indeed, writing
$A = A_x \mathbb i + A_y \mathbb j + A_z \mathbb k, \tag{2}$
$B = B_x \mathbb i + B_y \mathbb j + B_z \mathbb k, \tag{3}$
we have
$A \times B = (A_y B_z - A_z B_y) \mathbb i + (A_z B_x - A_x B_z) \mathbb j + (A_x B_y - A_y B_x) \mathbb k; \tag{4}$
thus, if $A \times B = 0$, 
$A_y B_z - A_z B_y = A_z B_x - A_x B_z = A_x B_y - A_y B_x = 0; \tag{5}$
assuming
$B_x, B_y, B_z \ne 0, \tag{6}$
which can always be arranged by a suitable choice of Cartesian vector basis on $\Bbb R^3$ (a fact I leave to my readers to verify for themselves), (5) gives rise to the ratio equations
$\dfrac{A_y}{B_y} = \dfrac{A_z}{B_z}, \tag{7}$
$\dfrac{A_x}{B_x} = \dfrac{A_z}{B_z}, \tag{8}$
and
$\dfrac{A_x}{B_x} = \dfrac{A_y}{B_y}.  \tag{9}$
Combining any two of (7)-(9) yields
$\dfrac{A_x}{B_x} = \dfrac{A_y}{B_y} = \dfrac{A_z}{B_z}; \tag{10}$
denoting the common ratio occurring in (10) by $c$, we see that
$A_x = cB_x, A_y = cB_y, A_z = cB_z, \tag{11}$
or 
$A = cB. \tag{12}$
Going the other way is even easier; starting from (12),
$A \times B = A \times cA = cA \times A = 0.   \tag{13}$
(1) may be had even more quickly provided we accept the standard $3$-vector identity
$X \times (Y \times Z) = (X \cdot Z) Y - (X \cdot Y) Z; \tag{14}$
with $A \times B = 0$ we have by (14)
$(B \cdot B)A - (B \cdot A) B = B \times (A \times B) = 0; \tag{15}$
since $B \ne 0$ implies $B \cdot B \ne 0$, (15) may be written
$A = \dfrac{A \cdot B}{B \cdot B}B; \tag{16}$
taking
$c = \dfrac{A \cdot B}{B \cdot B} \tag{17}$
yields (12). 
Applying these considerations to the problem at hand indicates we must take
$\vec k$ to be a scalar multiple of $\vec F$ to achieve the desired result, and that any such multiple of $\vec F$ will suffice.   The factors of $i$ or 
$e^{i(\vec k \cdot \vec r - \omega t)}$ do not affect this conclusion.