Animated Solution for Mathematics - Three Dimensional Geometry: Let the line $\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z-3}{-4}$ intersect the plane containing the lines $\frac{x-4}{1}=\frac{y+1}{-2}=\frac{z}{1}$ and $4ax-y+5z-7a=0=2x-5y-z-3,a\in \mathbb{R}$ at the point $P(\alpha ,\beta ,\gamma )$. Then the value of $\alpha +\beta +\gamma$ equals ______.

We are navigating the architecture of 3D space to find the intersection point between a line $L_1$ and a plane $M$. The plane $M$ is defined by its relationship to other objects: it contains a line $L_2$ and is formed by the intersection of two other planes, $P_1$ and $P_2$.

This is a classic JEE Advanced setup, designed to test your ability to synthesize multiple geometric constraints. Let us begin the derivation.

We define the plane $M$ using the family of planes equation. Given $P_1:4ax-y+5z-7a=0$ and $P_2:2x-5y-z-3=0$, any plane passing through their intersection is given by:

By grouping the terms, we obtain the general equation for plane $M$:

We have two unknowns, $a$ and $\lambda$, which we must determine using the properties of line $L_2$.

Our first condition arises because line $L_2$ lies entirely on plane $M$. Every point on $L_2:\frac{x-4}{1}=\frac{y+1}{-2}=\frac{z}{1}$ must satisfy the equation of $M$. Substituting the point $(4,-1,0)$ into the plane equation:

Expanding and simplifying this expression yields our first constraint:

The second condition arises because the direction vector of $L_2$, $\overrightarrow{d_2}=(1,-2,1)$, must be perpendicular to the normal vector of $M$, $\vec{n}=(4a+2\lambda ,-1-5\lambda ,5-\lambda )$. Setting their dot product to zero:

Expanding this gives $4a+2\lambda +2+10\lambda +5-\lambda =0$, which simplifies to:

Solving the system of equations $9a+10\lambda +1=0$ and $4a+11\lambda +7=0$ yields $\lambda =-1$ and $a=1$. Substituting these into our general equation, the plane $M$ simplifies to:

Now that we have the equation of plane $M$, we return to line $L_1:\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z-3}{-4}$. We express $L_1$ in parametric form:

Substituting these into the plane equation $x+2y+3z-2=0$:

Expanding this results in:

Plugging $t=2$ back into the parametric equations, we find the coordinates of the intersection point $P$:

The point of intersection is $P(17,0,-5)$. The sum of the coordinates $\alpha +\beta +\gamma$ is: