Step 1: Understanding the Question:
We need to convert a line given by the intersection of two planes (Cartesian equations) into its standard vector equation form $\vec{r} = \vec{a} + \lambda\vec{b}$.
Step 2: Key Formula or Approach:
Rearrange the Cartesian equations into the symmetric form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$. Then extract the position vector of a point on the line $\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$ and the direction vector $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$.
Step 3: Detailed Explanation:
The given equations are $y = 2$ and $4x - 3z + 5 = 0$.
Rewrite $y = 2$ as $y - 2 = 0$.
Rewrite the second equation to relate $x$ and $z$:
$4x = 3z - 5 \implies 4x = 3\left(z - \frac{5}{3}\right)$
Divide both sides by 12 (the LCM of 4 and 3) to put it into standard ratio form:
$\frac{x}{3} = \frac{z - \frac{5}{3}}{4}$
Combining both parts, the symmetric form of the line is:
$\frac{x - 0}{3} = \frac{y - 2}{0} = \frac{z - \frac{5}{3}}{4}$
From this, we identify a point on the line: $\left(0, 2, \frac{5}{3}\right)$.
The position vector of this point is $\vec{a} = 2\hat{j} + \frac{5}{3}\hat{k}$.
The direction ratios are $(3, 0, 4)$, so the direction vector is $\vec{b} = 3\hat{i} + 4\hat{k}$.
The vector equation is $\vec{r} = \vec{a} + \lambda\vec{b}$:
$\vec{r} = \left(2\hat{j} + \frac{5}{3}\hat{k}\right) + \lambda(3\hat{i} + 4\hat{k})$
Step 4: Final Answer:
The vector equation matches option (B).