Problem: Find the equation of the plane containing the line of intersection of the planes $x+y+z=0$ and $2x + y – 3z = 2$, and perpendicular to the plane $x – 2y – 5z = 17$.
Attempt at a solution: The direction vector of the line of intersection is given by the crossproduct of the two normals of the planes. The two normals are $(1,1,1)$ and $(2,1,-3)$. The crossproduct of these is \begin{align*} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 1 & – 3 \end{vmatrix} \end{align*} which gives $\hat{i}(-3-1) + \hat{j} (-3 – 2) + \hat{k}(1-2)$. Hence the direction vector of the line of intersection is \begin{align*} \mathbf{r} = -4 \hat{i} – 5 \hat{j} – \hat{k} \end{align*} Since our plane must also be perpendicular to the plane $x – 2y – 5z = 17$, it will be parallel to the normal of that plane, which is \begin{align*} \mathbf{n} = \hat{i} – 2 \hat{j} – 5 \hat{k} \end{align*} I'm not sure where to go from here, any help?
Edit: The direction vector is $\mathbf{r} = -4 \hat{i} + 5 \hat{j} – \hat{k}$. A point on the line of intersection (setting $x = 0$) is $(0,1/2,-1/2)$. A normal to our plane can be found by taking the crossproduct of $\mathbf{n}$ and our directionvector $\mathbf{r}$. The normal is \begin{align*} \mathbf{a} = 27 \hat{i} + 21 \hat{j} – 3 \hat{k}. \end{align*} Hence the equation of the plane is \begin{align*} 27(x – 0) + 21 (y-1/2) + 3 (z + 1/2) = 0.\end{align*}
Best Answer
Once you know that the plane contains the vector $r$ and the vector $n$, you can determine its normal vector $k$. It also must contain the intersection line, hence contain at least one point of that line. So you need to find a single point $P$ of that line. I suggest you do so by setting $x = 0$ and solving for $y$ and $z$. Once you've done that, the plane equation is
$$ (X - P) \cdot k = 0. $$
But as I noted in my comment, you've computed $r$ incorrectly, so you need to fix that.