Answer
A parametrization for the line is
$\left\{\begin{array}{l}
x=1+2t\\
y=-3+5t\\
z=1+4t
\end{array}\right., \quad -\infty \lt t \lt \infty$
(Other answers are possible.)
Work Step by Step
The line belongs to both planes, so it is perpendicular to both normal vectors (we use 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), $\left\{\begin{array}{l}
x=x_{0}+v_{1}t\\
y=y_{0}+v_{2}t\\
z=z_{0}+v_{3}t
\end{array}\right., \quad -\infty \lt t \lt \infty$
${\bf n_{1}}\times{\bf n_{2}}=\left|\begin{array}{lll}
{\bf i} & {\bf j} & {\bf k}\\
5 & -2 & 0\\
0 & 4 & -5
\end{array}\right|=\langle (10-0), -(-25-0), (20-0) \rangle$
$=\langle 10, 25, 20 \rangle=5\langle 2, 5, 4 \rangle$
We take ${\bf v}=\langle 2, 5, 4 \rangle$
Take for x and y a pair of numbers that satisfy the first plane equation
$5x=11+2y$
$x=1, y=-3$
Now, substitute $y=-3$ into the second equation:
$4(-3)-5z=-17$
$-5z=-5$
$z=1...\qquad\Rightarrow\quad P(1,-3,1)$ lies on the line.
A parametrization for the line is
$\left\{\begin{array}{l}
x=1+2t\\
y=-3+5t\\
z=1+4t
\end{array}\right., \quad -\infty \lt t \lt \infty$
