Equation of plane perpendicular to given plane Find the equation of the plane which contains the line of intersection of the planes $x+2y+3z-4=0$ and $2x+y-z+5=0$ and which is perpendicular to the plane $5x+3y-6z+8=0$
By setting $z=0$ I found a point $(-\frac{14}{3}, \frac{13}{3}, 0)$ but not able to obtain normal vector of the plane. Would finding complete equation of line of intersection be more helpful?
 A: In answer to your last question, yes: because our required plane, which we shall call $P$, must contain a vector, $\mathbf{l}$ say, parallel to our line of intersection and also the vector, $\mathbf{n} := (5,3,-6)$, normal to the plane $5x+3y-6y+8=0$.
Vector $\mathbf{l}$ is perpendicular to the normals: $(1,2,3)$ and $(2,1,-1)$, which means it equals their cross product:
$$\mathbf{l} = \begin{vmatrix}
        \mathbf{i} & \mathbf{j} & \mathbf{k} \\
        1 & 2 & 3 \\
        2 & 1 & -1 \\
        \end{vmatrix} = (-5,7,-3).
$$
Because plane $P$ contains both $\mathbf{l}$ and $\mathbf{n}$, its normal, $\mathbf{n_p}$ say, equals their cross product:
$$\mathbf{n_p} = \begin{vmatrix}
        \mathbf{i} & \mathbf{j} & \mathbf{k} \\
        -5 & 7 & -3 \\
        5 & 3 & -6 \\
        \end{vmatrix} = (-33,-45,-50).
$$
Multiplying it by $-1$ for convenience, we have $\mathbf{n_p} = (33,45,50)$. So $P$ has equation $33x+45y+50z=c\;$ for some value $c$. Our point $\left(\dfrac{-14}{3},\dfrac{13}{3},0\right)$ lies on $P$ so we substitute it into this equation to find $c$:
$$c = 33\times \dfrac{-14}{3} + 45\times \dfrac{13}{3}= 41.$$
So our plane $P$ is:
$$33x+45y+50z=41.$$
