Finding the bundle of planes through a point $P$, which doesn't intersect a line, in $\mathbb{R^3}$

Question

Consider in $\mathbb{R^3}$ the line $r = \begin{cases} \ x-y-z=2 \\[2ex] 2x-5y-8z=10\end{cases}$ , and the point $P=(1, 0, -2)$.

Determine the equation of a plane $\beta$ through $P$ but that doesn't intersect the line $r$. Then determine the equation of the bundle of planes through $P$ but not $r$.

I found $\beta$ by calculating a vector $w$ normal to the direction vector of the line $v_r =(-2, -3, 1)$, so $w=(1, 1, 5)$. Replacing in the general equation of a plane: $x+y+5z+d=0$ and then imposing to pass through $P$: $d=9 \to \beta:x+y+5x=-9$

How to find the bundle of planes which doesn't intersect $r$? Thank you

Answer 1

We need to find the bundle of normal vectors $\vec w$ to the direction vector of the line $r$ such that

$$(-2, -3, 1)\cdot(a, b, c)=0$$

and excluding the two values for which the plane $\beta$ contains the line $r$.