Answer
$\left\{\begin{array}{l}
x=4\\
y=1+2t\\
z=t
\end{array}\right., \quad -\infty \lt t \lt \infty$
(Other answers are possible.)
Work Step by Step
The line belongs to both planes, so it is perpendicular to both normal vectors (we use the cross product to find a direction vector for the line).
We find a point P that belongs to both planes and
with $P(x_{0},y_{0},z_{0})$ and ${\bf v}=\langle v_{1},v_{2},v_{3}\rangle$, the standard parametrization is given by formula (3), $\left\{\begin{array}{l}
x=x_{0}+v_{1}t\\
y=y_{0}+v_{2}t\\
z=z_{0}+v_{3}t
\end{array}\right., \quad -\infty \lt t \lt \infty$
${\bf n_{1}}\times{\bf n_{2}}=\left|\begin{array}{lll}
{\bf i} & {\bf j} & {\bf k}\\
1 & -2 & 4\\
1 & 1 & -2
\end{array}\right|=\langle (4-4), -(-2-4), (1+2) \rangle$
$\langle 0, 6, 3 \rangle=3\langle 0, 2, 1 \rangle$
We take ${\bf v}=\langle 0, 2, 1 \rangle$
Now, to find a point P.
Subtracting the plane equations, we arrive at
$-3y+6z=-3\qquad/\div(-3)$
$y-2z=1$
Take $z=0.$ It follows that $y=1$. Insert both of these into the first equation:
$x-2+0=2\Rightarrow x=4$
the point $P(4,1,0)$ lies in both planes -- that is, on our line.
We already found ${\bf v}=\langle 0, 2, 1 \rangle$.
A parametrization for the line is
$\left\{\begin{array}{l}
x=4\\
y=1+2t\\
z=t
\end{array}\right., \quad -\infty \lt t \lt \infty$
