The line $\dfrac{x-2}{3}=\dfrac{y-3}{4}=\dfrac{z-4}{5}$ is parallel to the plane

The direction vector of the line $\dfrac{x-2}{3} = \dfrac{y-3}{4} = \dfrac{z-4}{5}$ is $\vec{v} = 3\hat{i} + 4\hat{j} + 5\hat{k}$.

A line is parallel to a plane if its direction vector $\vec{v}$ is perpendicular to the normal vector $\vec{n}$ of the plane, which implies the dot product $\vec{v} \cdot \vec{n} = 0$.

Checking the options:

For option (1), the normal vector is $\vec{n} = 3\hat{i} + 4\hat{j} + 5\hat{k}$. The dot product is $\vec{v} \cdot \vec{n} = (3)(3) + (4)(4) + (5)(5) = 9 + 16 + 25 = 50 \neq 0$.

For option (2), the normal vector is $\vec{n} = 2\hat{i} + 3\hat{j} + 4\hat{k}$. The dot product is $\vec{v} \cdot \vec{n} = (3)(2) + (4)(3) + (5)(4) = 6 + 12 + 20 = 38 \neq 0$.

For option (3), the normal vector is $\vec{n} = 1\hat{i} + 1\hat{j} - 1\hat{k}$. The dot product is $\vec{v} \cdot \vec{n} = (3)(1) + (4)(1) + (5)(-1) = 3 + 4 - 5 = 2 \neq 0$.

For option (4), the normal vector is $\vec{n} = 2\hat{i} + 1\hat{j} - 2\hat{k}$. The dot product is $\vec{v} \cdot \vec{n} = (3)(2) + (4)(1) + (5)(-2) = 6 + 4 - 10 = 0$.

Since the dot product is zero, the line is parallel to the plane $2x + y - 2z = 0$.

Answer: $2 x+y-2 z=0$