How to find a line that intersects two planes? We have the two planes as $x=0$ and $z=-5$.

Hint: In this question we have been given with the equations of two planes and we have to find the line which intersects the two planes. We will find the line by considering the vector form of the equation of the line which is written as $\left( x,y,z \right)=\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)+t\vec{v}$. And then write the equation of the planes in orthogonal form and find the cross product. We will then substitute the values of $x$ and $y$, and solve for the value of the line in vector form.

Complete step by step solution:
We know that the vector form of a line is:
$\left( x,y,z \right)=\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)+t\vec{v}\to \left( 1 \right)$
Now the orthogonal vector for plane $x=0$ will be:
$1\hat{i}+0\hat{j}+0\hat{k}$
And the orthogonal vector for plane $x=0$ will be:
$0\hat{i}+0\hat{j}+1\hat{k}$
Since the vector intersecting the two planes will be orthogonal to both the planes, we can compute the vector using the cross product as:
$\Rightarrow \vec{v}=\left( 1\hat{i}+0\hat{j}+0\hat{k} \right)\times \left( 0\hat{i}+0\hat{j}+1\hat{k} \right)$
On cross multiplying, we get:
$\Rightarrow \vec{v}=0\hat{i}-1\hat{j}+0\hat{k}$
In the general form, it can be written as:
$\Rightarrow \vec{v}=\left( 0,-1,0 \right)$
Substituting the vector in equation $\left( 1 \right)$, we get:
$\Rightarrow \left( x,y,z \right)=\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)+t\left( 0,-1,0 \right)$
Now for the line to be on the plane $x=0$, $x$ must always be $0$ therefore, it can be written as:
$\Rightarrow \left( 0,y,z \right)=\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)+t\left( 0,-1,0 \right)$
Since the value of $x=0$, it will force ${{x}_{1}}=0$ therefore, it can be written as:
$\Rightarrow \left( 0,y,z \right)=\left( 0,{{y}_{1}},{{z}_{1}} \right)+t\left( 0,-1,0 \right)$
Now for the line to be on the plane $z=-5$, $x$ must always be $-5$ therefore, it can be written as:
$\Rightarrow \left( 0,y,-5 \right)=\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)+t\left( 0,-1,0 \right)$
Since the value of $z=-5$, it will force ${{z}_{1}}=-5$ therefore, it can be written as:
$\Rightarrow \left( 0,y,z \right)=\left( 0,{{y}_{1}},-5 \right)+t\left( 0,-1,0 \right)$
This implies that the value of $y$ is the only value which is changing and since the domain of $t$ is real numbers, the value of ${{y}_{1}}$ does not matter which means we can choose any value. On choosing ${{y}_{1}}=0$, we get:
$\Rightarrow \left( 0,y,z \right)=\left( 0,0,-5 \right)+t\left( 0,-1,0 \right)$
For the same reason we can also change the value of $-1$ to $1$, therefore, we get:
$\Rightarrow \left( 0,y,z \right)=\left( 0,0,-5 \right)+t\left( 0,1,0 \right)$, which gives us the required vector equation of the line.
Therefore, the parametric equations are:
$x=0,y=t,z=-5$
And the symmetric form is $y$, with the restrictions $x=0$ and $z=-5$.

Note: In this question we are dealing with vectors in three-dimensions. There are three coordinates for a point in three-dimensions which are $x,y$ and $z$ respectively. it is to be remembered that for a given point $A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and the directions of the ratio lines $a,b$ and $c$, the parametric equation of the line is:
$x={{x}_{1}}+ka$,$y={{y}_{1}}+kb$ and $z={{z}_{1}}+kc$.