Find the Cartesian equation of the plan parallel to j and passes through the intersection of the planes described by the equations x + 2y + 3z = 4, and 2x + y + z = 2.
I was able to get the parametric equation of the intersection line: (0, 2, 0) + t(1, -5, 3), but in this regard, there would not be any plane containing this line that is parallel to j, since this line itself is not parallel to j, correct? Or am I missing something?
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if the two given planes are $f=0$ and $g=0$ then any other plane sharing their line of intersection has equation: $$ \lambda f + g = 0 $$ this gives: $$ (\lambda+2)x +(2\lambda+1)y +(3 \lambda+1) = 4\lambda +2 $$ to be parallel to the Y-axis we require the coefficient of $y$ to be zero, hence $$ \lambda = -\frac12 $$ giving: $$ 3x - z = 0 $$
It's true that the line is not parallel to $\textbf{j}$, but that's fine. There does exist a plane that both contains your intersection line and is parallel to $\textbf{j}$. To find this plane, we need a normal vector $\textbf{n}$ and an initial point. For the initial point, we can use the same initial point as the intersection line: $(0, 2, 0)$.
For the normal vector, notice that it must be perpendicular to any vector in the plane or any vector parallel to the plane. So we may take the cross-product $\textbf{n} = \textbf{j} \times \textbf{v}$, where $\textbf{v} = (1, -5, 3)$ is the direction vector of the intersection line.