Here is one way:
Line $p_1$ is defined by $(1,2,3)x+(0,-1,2)$, while $p_2$ is defined by $(1,-1,4)x+(0,1,-1)$.
So the direction vector for $p1$ is $(1,2,3)$ and the direction vector for $p2$ is $(1,-1,4)$. A quick way to find a vector perpendicular to both of them is to calculate their cross product, which is $(11,-1,-3)$.
Therefore any plane that is perpendicular to the vector $(11,-1,-3)$ will be parallel to or coincident with your two given lines. The equation of such a plane is
$$11x-y-3z=c$$
for some constant $c$.
If the plane is equidistant from the two lines, then the value of $c$ will be equidistant from the values of that expression when a point on each line is substituted. The value of $11x-y-3z$ for point $(0,-1,2)$ on line $p_1$ is $-5$, while the value for point $(0,1,-1)$ on line $p_2$ is $2$.
Therefore we want $c$ to be the average of $-5$ and $2$, which is $-\frac 32$. Your desired plane is
$$11x-y-3z=-\frac 32$$
Alternatively, if you don't want to fuss over expanding the determinants as I told you to try, you can use this argument. Since $V:=u_1\cdot\left(u_2\times u_3\right)=u_2\cdot\left(u_3\times u_1\right)=u_3\cdot\left(u_1\times u_2\right) \neq 0$, the vectors $u_1$, $u_2$, and $u_3$ are linearly independent. Therefore, the solution $r\in\mathbb{R}^3$ to the system of equations $r\cdot u_i-d_i=0$ for $i=1,2,3$ is unique. Now, if $$r=\frac{d_1\left(u_2\times u_3\right)+d_2\left(u_3\times u_1\right)+d_3\left(u_1\times u_2\right)}{V}\,,$$ then
$$r\cdot u_1=\frac{d_1\left(u_2\times u_3\right)\cdot u_1}{V}=\frac{d_1 V}{V}=d_1$$
since $\left(u_3\times u_1\right)\perp u_1$ and $\left(u_1\times u_2\right)\perp u_1$. Likewise, $r\cdot u_2=d_2$ and $r\cdot u_3=d_3$. Hence, this formula for $r$ gives the unique solution to the system of equations.
There is still another solution. You can refer to this vector identity:
$$
\left(\mathbf{A}\cdot\left(\mathbf{B}\times\mathbf{C}\right)\right)\mathbf{D}=\left(\mathbf{A}\cdot\mathbf{D}\right)\left(\mathbf{B}\times\mathbf{C}\right)+\left(\mathbf{B}\cdot\mathbf{D}\right)\left(\mathbf{C}\times\mathbf{A}\right)+\left(\mathbf{C}\cdot\mathbf{D}\right)\left(\mathbf{A}\times\mathbf{B}\right) \,.$$
For reference, see https://en.wikipedia.org/wiki/Vector_calculus_identities.
In general, the solution $r\in\mathbb{R}^n$ to $r\cdot u_i-d_i=0$ for $i=1,2,\ldots,n$ is given by
$$r=\frac{\sum_{i=1}^n\,(-1)^{i-1}d_i\left(u_1\wedge u_2\wedge \ldots \wedge u_{i-1} \wedge u_{i+1}\wedge \ldots \wedge u_{n-1}\wedge u_n\right)}{u_1\wedge u_2 \wedge \ldots \wedge u_{n}}\,,$$
where we identify $\bigwedge^{n-k}\,\mathbb{R}^n$ as $\bigwedge^{k}\,\mathbb{R}^n$.
Best Answer
Define the matrix $$M= {\begin{bmatrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{bmatrix}} $$ and consider its effect on points on the surface of the parallelepiped:
Thus the first pair or parallel planes transform under $M$ to two planes perpendicular to the $x$-axis and separated by a distance $|p_1-q_1|$.
Likewise the second pair of parallel planes transform under $M$ to two planes perpendicular to the $y$-axis and separated by a distance $|p_2-q_2|$; and the third pair of parallel planes transform under $M$ to two planes perpendicular to the $z$-axis and separated by a distance $|p_3-q_3|$.
So the parallelepiped transforms into a rectangular cuboid, of volume $V=\prod_{i=1}^{3} |p_i - q_i|$. And the inverse matrix $M^{-1}$ trasforms a solid of volume $V$ into the original parallelepiped.
Hence the volume of the parallelepiped is $|V\det(M^{-1})|=\left|\dfrac{V}{\det M}\right|$.