直线 $\ell$ 的方程为 $\dfrac{x-a}{2}=y-2=\dfrac{1-2z}{b}$，其中 $a$ 和 $b$ 为实常数。平面 $\pi$ 的方程为 $3x-y+4z=10$。

1. Express the Cartesian equation of the line $\ell$ in parametric form.

Let $\dfrac{x-a}{2} = y-2 = \dfrac{1-2z}{b} = \lambda$.

Expressing $x$, $y$, and $z$ in terms of $\lambda$:
$x = a + 2\lambda$
$y = 2 + \lambda$
$1 - 2z = b\lambda \implies z = \dfrac{1}{2} - \dfrac{b}{2}\lambda$

2. Substitute the parametric coordinates into the equation of plane $\pi$.

Substitute $x$, $y$, and $z$ into the plane equation $3x - y + 4z = 10$:
$3(a + 2\lambda) - (2 + \lambda) + 4\left(\dfrac{1}{2} - \dfrac{b}{2}\lambda\right) = 10$

3. Expand and simplify the equation to group the $\lambda$ terms.

$3a + 6\lambda - 2 - \lambda + 2 - 2b\lambda = 10$
$(6 - 1 - 2b)\lambda + 3a = 10$
$(5 - 2b)\lambda = 10 - 3a$

4. Apply the condition for no intersection to prove the results.

For the line and the plane to have no intersection, they must be parallel but not coincident. This means the resulting linear equation must have no solution for $\lambda$.

This occurs when the coefficient of $\lambda$ is equal to $0$ and the constant term on the RHS is non-zero:

1. $5 - 2b = 0 \implies b = \dfrac{5}{2}$
2. $10 - 3a \ne 0 \implies a \ne \dfrac{10}{3}$

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1. Express the Cartesian equation of the line in parametric form.

Substitute $a=1$ and $b=3$ into the equation of $\ell$ and set it to a parameter $\lambda$:
$\frac{x-1}{2} = y-2 = \frac{1-2z}{3} = \lambda$

Expressing $x$, $y$, and $z$ in terms of $\lambda$:
$x = 1 + 2\lambda$
$y = 2 + \lambda$
$1 - 2z = 3\lambda \implies z = \frac{1}{2} - \frac{3}{2}\lambda$

2. Substitute the parametric equations into the equation of the plane to find the parameter value.

Substitute the expressions for $x$, $y$, and $z$ into the equation of the plane $\pi$, $3x - y + 4z = 10$:
$3(1 + 2\lambda) - (2 + \lambda) + 4\left(\frac{1}{2} - \frac{3}{2}\lambda\right) = 10$
$3 + 6\lambda - 2 - \lambda + 2 - 6\lambda = 10$
$3 - \lambda = 10$
$\lambda = -7$

3. Calculate the coordinates of the point of intersection.

Substitute $\lambda = -7$ back into the parametric equations:
$x = 1 + 2(-7) = -13$
$y = 2 + (-7) = -5$
$z = \frac{1}{2} - \frac{3}{2}(-7) = \frac{1}{2} + \frac{21}{2} = 11$

The point of intersection is $(-13, -5, 11)$.

1. Identify a point on the line $\ell$.

By setting the parameter of the line equation $\frac{x-1}{2} = y-2 = \frac{1-2z}{5/2}$ to $0$, we get:
$x - 1 = 0 \implies x = 1$
$y - 2 = 0 \implies y = 2$
$1 - 2z = 0 \implies z = \frac{1}{2}$

A point on the line $\ell$ is $A\left(1, 2, \frac{1}{2}\right)$.

2. Calculate the perpendicular distance from the point to the plane $\pi$.

Since the line is parallel to the plane, the distance between them is the perpendicular distance from point $A$ to the plane $3x - y + 4z = 10$.

Using the point-to-plane distance formula $d = \frac{|n_1x_0 + n_2y_0 + n_3z_0 - D|}{\sqrt{n_1^2 + n_2^2 + n_3^2}}$:
$\text{Distance} = \frac{\left| 3(1) - 1(2) + 4\left(\frac{1}{2}\right) - 10 \right|}{\sqrt{3^2 + (-1)^2 + 4^2}}$
$d = \frac{| 3 - 2 + 2 - 10 |}{\sqrt{9 + 1 + 16}}$
$d = \frac{| -7 |}{\sqrt{26}} = \frac{7}{\sqrt{26}}$

1. Determine whether the origin and the line lie on the same side of the plane $\pi$.

Let the plane $\pi$ be represented by $\mathrm{f}(x, y, z) = 3x - y + 4z - 10$.
Substituting the coordinates of the origin $O(0, 0, 0)$:
$\mathrm{f}(0,0,0) = 3(0) - 0 + 4(0) - 10 = -10$
From the equation of the line $\ell$, we can identify a point $A$ on the line by setting the numerators to zero:
$x-1=0 \implies x=1, \quad y-2=0 \implies y=2, \quad 1-2z=0 \implies z=\frac{1}{2}$
Substituting point $A\left(1, 2, \frac{1}{2}\right)$ into $\mathrm{f}$:
$\mathrm{f}\left(1, 2, \frac{1}{2}\right) = 3(1) - 2 + 4\left(\frac{1}{2}\right) - 10 = -7$
Since $\mathrm{f}(0,0,0) < 0$ and $\mathrm{f}\left(1, 2, \frac{1}{2}\right) < 0$ (both evaluate to a negative value), the point $A$ (and thus the entire line $\ell$) and the origin $O$ lie on the same side of the plane $\pi$.

2. State the equation of the other plane equidistant from line $\ell$ and parallel to plane $\pi$.

Since $\pi$ is parallel to $\ell$, the plane containing $\ell$ and parallel to $\pi$ has the equation $3x - y + 4z = c$.
Using point $A\left(1, 2, \frac{1}{2}\right)$ on $\ell$:
$c = 3(1) - 2 + 4\left(\frac{1}{2}\right) = 3$
The plane containing $\ell$ is $3x - y + 4z = 3$.
Let the required other plane be $3x - y + 4z = d$.
For this plane and $\pi$ ($3x - y + 4z = 10$) to be equidistant from $\ell$, the plane containing $\ell$ must lie exactly halfway between them:
$\frac{10 + d}{2} = 3$
$10 + d = 6 \implies d = -4$
Therefore, the equation of the other plane is $3x - y + 4z = -4$.