Step 1: Understand the line.
The line is given by two plane equations $y=2$ and $4x-3z+5=0$. The line is where these two planes cross. We want it as $\vec r=\vec a+\lambda\vec b$.
Step 2: Aim for symmetric form.
We want $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$. From this we read a point $(x_1,y_1,z_1)$ and a direction $(a,b,c)$.
Step 3: Handle the $y$ part.
Since $y=2$ is fixed, write it as $y-2=0$. The line does not move in the $y$ direction, so its $y$-direction number is $0$.
Step 4: Handle the $x$ and $z$ part.
From $4x-3z+5=0$ we get $4x=3z-5=3\!\left(z-\frac{5}{3}\right)$. Divide by $12$: \[ \frac{x}{3}=\frac{z-\frac{5}{3}}{4}. \]
Step 5: Write the full symmetric form.
\[ \frac{x-0}{3}=\frac{y-2}{0}=\frac{z-\frac{5}{3}}{4}. \]
Step 6: Read the point and direction.
A point on the line is $\left(0,2,\frac{5}{3}\right)$, so $\vec a=2\hat j+\frac{5}{3}\hat k$. The direction numbers are $(3,0,4)$, so $\vec b=3\hat i+4\hat k$.
Step 7: Build the vector equation.
\[ \vec r=\left(2\hat j+\frac{5}{3}\hat k\right)+\lambda\left(3\hat i+4\hat k\right). \] This is option (2).
\[ \boxed{\vec r=\left(2\hat j+\tfrac{5}{3}\hat k\right)+\lambda(3\hat i+4\hat k)} \]