We have to find the equation of the plane through the line
$\frac{x - \alpha}{l} = \frac{y - \beta}{m} = \frac{z - \gamma}{n} $
My method was to use $a(x-\alpha\ ) + b(x-\beta\ ) + c(z-\gamma\ )=0$, where a,b,c are the direction ratios of the normal to the plane, and solve it further.
But, the solution provided mentions an alternative method, saying that the equation of the plane can be expressed as :- $m(x-\alpha\ ) - l(y-\beta\ ) + \lambda\ [n(y-\beta\ ) -m(z-\gamma\ )]=0$
I can understand that these are the equations of 2 planes through which that line passes, but only after cross checking the angle between these plane and the line being zero.
How can we express the plane like that?
Since $\frac{x-\alpha}{l} = \frac{y-\beta}{m}$, this means $m(x-\alpha)-l(y-\beta)=0$. This is an equation for a plane that the line is contained in. Analogously, from $\frac{y-\beta}{m}=\frac{z-\gamma}{n}$ we get $n(y-\beta)-m(z-\gamma)=0$ is another plane.
Taking a linear combination of these two equations we get:
$$\mu(m(x-\alpha)-l(y-\beta)) + \lambda(n(y-\beta)-m(z-\gamma)) = 0$$
Note that this equation still holds for any point on the line. Now, if $\mu=\lambda=0$ the equation doesn't represent a plane, so at least one of them must be different to zero. We arbitrarily decide that $\mu$ is not zero, and thus can divide the equation by $\mu$ to get the equation in the solution.
We can check that this includes all planes by looking at the normal vectors of the planes. Notice that making linear combinations of $(m, -l, 0)$ and $(0, n, -m)$ (the normal vectors of the first two planes) we can reach any vector perpendicular to the line, since they are linearly independent ($m,n,l \neq 0$) and each normal vector would correspond to a point on a plane perpendicular to the line. Finally, for any vector perpendicular to the line, the vector $(m, -l+\lambda n, -\lambda m)$ (the normal vector of the solution) is proportional to it for a specific $\lambda$.