Derive Equation of Plane passing through the intersection of two planes 
How to derive the equation of the plane passing through the intersection of two given planes.

Lets say we have given two planes
$$
\pi_1 : \vec{r}.\hat{n}_1=d_1\\
\pi_2 : \vec{r}.\hat{n}_2=d_2\\
$$
where $\hat{n}_1, \hat{n}_2$ : unit vectors normal to the planes $\pi_1, \pi_2$ and $d_1, d_2$ : perpendicular distances from the origin.
The position vector of any point on the line of intersection must satisfy both the equations.
So far good, I understand this. But, from here how do I prove that any plane passing through the intersection of the planes is:
$$\boxed{
\pi_3 : \vec{r}.(\hat{n}_1+\lambda \hat{n}_2)=d_1+\lambda d_2\\
\qquad\qquad\quad\text{OR}\\
\pi_3 : \vec{r}.(\alpha\hat{n}_1+\beta \hat{n}_2)=\alpha d_1+\beta d_2
}$$
My Understanding
From the figure 

and parallelogram law of vector addition, the normal of the plane passing through the intersection of $\pi_1$ and $\pi_2$ will be some linear combination of $n_1$ and $n_2$. ie, $\hat{n}_3=\alpha\hat{n}_1+\beta\hat{n}_2$. And any point should satisfy equations of $\pi_1$ and $\pi_2$. Thus,
$$
\vec{r}.\hat{n}_3=D \implies \vec{r}.(\alpha\hat{n}_1+\beta\hat{n}_2)=\alpha\vec{r}.\hat{n}_1+\beta\vec{r}.\hat{n}_2=\alpha d_1+\beta d_2=D\\\color{red}{
\implies \vec{r}.(\alpha\hat{n}_1+\beta\hat{n}_2)=\alpha d_1+\beta d_2}
$$
Is it the right explanation of the derivation ?
 A: All the points that satisfy the first (second) equation belong to the first (second) plane.
You are free to form a linear combination of two equations, for instance with the coefficients $1$ and $\lambda$.
$$\vec r\vec n_1+\lambda\,\vec r\vec n_2=d_1+\lambda\,d_2,$$ which can be written
$$\vec r\,(\vec n_1+\lambda\,\vec n_2)=d_1+\lambda\,d_2.$$
The resulting equation has the shape of a plane equation. Furthermore, any point that satisfies the two given equations will also verify the combined one. So the new equation describes a plane that contains the intersection of the two given planes.
In fact, if you vary $\lambda$, you get an infinity of different planes (among which the plane $1$ for $\lambda=0$; this parameterization does not allow to include the plane $2$).
If you admit that all the possible solution planes are defined by the line of intersection and a point outside this line, then you can plug the point in the combined equation and draw the value of $\lambda$. If the point is outside the plane $2$, there is a solution. Hence the equation describes all possible planes (but $2$).
A: I think your figure gives a good intuition why the formulas work,
provided that the planes $\pi_1$ and $\pi_2$ are distinct and not parallel.
(If $\pi_1$ and $\pi_2$ are identical or parallel to each other, the formulas for $\pi_3$ give a plane parallel to $\pi_1$ and $\pi_2$.)
Assuming $\pi_1$ and $\pi_2$ are distinct and not parallel,
take a plane $\pi_\perp$ perpendicular to the line of intersection;
any plane through the line of intersection must have a normal vector parallel to that plane.
The two normal vectors in the figure are both parallel to $\pi_\perp$,
and they span the two-dimensional space of vectors consisting of all vectors parallel to $\pi_\perp$.
In other words, the set of non-zero linear combinations of the two normals is exactly the set of normals to all planes through the line of intersection.
That gives you the formula
$$\pi_3 : \vec{r}.(\alpha\hat{n}_1+\beta \hat{n}_2)=\alpha d_1+\beta d_2,$$
and the other formula can be derived from that as long as $\alpha \neq 0$
(that is, it can represent every such plane except the given plane $\pi_2$).
A: Since $\vec t$ is on the intersecting line, $\vec t$ satisfies $\pi_1$ and $\pi_2$. So it also satisfies a linear combination of $\pi_1$ and $\pi_2$.
$$\begin{cases}\vec{t}\cdot \hat{n}_1=d_1\\
\vec{t}\cdot\hat{n}_2=d_2\end{cases}$$ If you multiply $\pi_2$ with a real $\lambda$, and add them together, you'll have $$\vec{t}\cdot \hat{n}_1+\lambda(\vec{t}\cdot\hat{n}_2)=d_1+\lambda d_2$$ and thus $$\vec{t}\cdot (\hat{n}_1+\lambda \hat{n}_2)=d_1+\lambda d_2.$$ You can of course do $\alpha\pi_1+\beta\pi_2$, but we want to ensure $\alpha,\beta$ are not all zero: say taking $\alpha=1,\beta=\lambda\in\mathbb{R}$.
A: Given two distinct intersecting planes
$$
\Pi_1\to a_1x+b_1y+c_1z+d_1=0\\
\Pi_2\to a_2x+b_2y+c_2z+d_2=0
$$
and now considering $\lambda_i \ne 0$,  given
$$
\Pi'_1\to \lambda_1\left(a_1x+b_1y+c_1z+d_1\right)=0\\
\Pi'_2\to \lambda_2\left(a_2x+b_2y+c_2z+d_2\right)=0
$$
The solution set for $\Pi_1\cap\Pi_2$ and for $\Pi'_1\cap\Pi'_2$ are identical. This solution set corresponds to the intersection line.
Adding $\Pi'_1$ and $\Pi'_2$ we obtain
$$
\Pi_{\lambda} \to (\lambda_1 a_1+\lambda_2 a_2)x+(\lambda_1 b_1+\lambda_2 b_2) y+(\lambda_1 c_1+\lambda_2c_2)z +\lambda_1d_1+\lambda_2 d_2 = 0
$$
now $\Pi_{\lambda}\cap\Pi_1 = \Pi_{\lambda}\cap\Pi_2$ hence by construction $\Pi_{\lambda}$ contains any plane which has common intersection with $\Pi_1$ and $\Pi_2$ 
NOTE
making $\lambda_1 = 1$ and $\lambda_2 = \lambda$ we have the proposed situation.
