Three planes have the equation $$ \pi_1:x+2y-3z=-4$$ $$ \pi_2:2x+y-4z=3$$ $$ \pi_3:x+2y-3z+4+\lambda(2x+y-4z-3)=0$$
Find the equation of the plane $\pi_3$ which passes through the line on which $\pi_1$ and $\pi_2$ meet and contains the point whose position vector is $\alpha k$
I first calculated the vector equation of the line on which the two planes meet. Using the position vector of the line and the $\alpha k$, I calculated a vector that $\pi_3$ contains. Cross product of line direction vector and obtained vector gives the normal to the plane. The dot product of the normal and the $\alpha k$ position vector completes the equation of the plane.
I obtained $$-(2\alpha +5)x+(5\alpha +2)y+7z=7a$$
The given answer is $$(-2\alpha +11)x+(5\alpha +10)y-25z=-25a$$
Can somebody please check if these answers are same? If no, then maybe explain the flaw in my strategy?
P.S. I am aware of the second method that involves directly putting the position vector in the $\pi_3$ plane.
Working
Line equation:$2/3 i +5/3 j+(5i+2j+3k)t$
Normal Obtained: $(5,2,3)\times(2/3,5/3,-\alpha)=-(2\alpha +5)i+(5\alpha +2)j+7k$
Dot product of normal and position vector :$7a$
If some answers seems doubtful, upon request, I will provide detailed working.