Animated Solution for Mathematics - Three Dimensional Geometry: The equation of the plane passing through the point $(1,2,-3)$ and perpendicular to the planes $3x+y-2z=5$ and $2x-5y-z=7$, is:

Finding Your Plane

Welcome, fellow explorer of the mathematical universe. Today, we are not just solving a problem; we are performing an act of architectural creation in three-dimensional space.

We are tasked with finding the equation of a plane that passes through a specific point, $A(1,2,-3)$, while standing perfectly perpendicular to two other planes, $P_1$ and $P_2$.

Imagine standing in a room where two walls meet at an angle. You are holding a sheet of glass (our target plane) and you need to orient it so that it is perfectly perpendicular to both walls. How do we define that orientation?

The Normal Vector - The Plane's Compass

In the language of vectors, a plane is defined by its normal vector—a vector that points straight out of the surface at a $90^{\circ }$ angle. Think of it as the plane's compass.

If we know the normal vector, we know the plane's tilt. For our given planes, $P_1:3x+y-2z=5$ and $P_2:2x-5y-z=7$, the normal vectors are hidden in plain sight.

They are simply the coefficients of $x,y,$ and $z$. Thus, for $P_1$, we have ${\vec{n}}_1=3\hat{i}+\hat{j}-2\hat{k}$, and for $P_2$, we have ${\vec{n}}_2=2\hat{i}-5\hat{j}-\hat{k}$. These two vectors are the keys to our kingdom.

The Cross Product - The Architect's Secret

Here is where the magic happens. We need our new plane to be perpendicular to both $P_1$ and $P_2$.

Geometrically, this means the normal vector of our new plane, let's call it $\vec{n}$, must be perpendicular to both ${\vec{n}}_1$ and ${\vec{n}}_2$. There is only one operation in our vector toolkit that produces a vector perpendicular to two others: the cross product.

By calculating $\vec{n}={\vec{n}}_1\times {\vec{n}}_2$, we are essentially finding the direction that is orthogonal to the 'tilt' of both existing planes.

The Determinant - A Dance of Numbers

To compute the cross product, we set up a determinant. This is where precision is paramount. A single sign error here can lead us astray.

We arrange our unit vectors $\hat{i},\hat{j},\hat{k}$ in the top row, followed by the components of ${\vec{n}}_1$ and ${\vec{n}}_2$ in the subsequent rows:

Expanding this, we calculate the $\hat{i}$ component: $(1)(-1)-(-5)(-2)=-1-10=-11$.

For the $\hat{j}$ component, we remember the alternating sign rule: $-((3)(-1)-(2)(-2))=-(-3+4)=-1$.

Finally, for the $\hat{k}$ component: $(3)(-5)-(2)(1)=-15-2=-17$. Our resultant normal vector is $\vec{n}=-11\hat{i}-\hat{j}-17\hat{k}$.

The Final Assembly

Now that we have our normal vector $\vec{n}=(-11,-1,-17)$, we can simplify our life by using $\vec{n}=(11,1,17)$ (multiplying by $-1$ does not change the plane's orientation).

We use the point-normal form: $a(x-x_1)+b(y-y_1)+c(z-z_1)=0$. Plugging in our point $A(1,2,-3)$ and our normal vector $(11,1,17)$, we get:

Expanding this carefully: $11x-11+y-2+17z+51=0$.

Grouping the terms, we arrive at the final equation: $11x+y+17z+38=0$.

Look at what you have achieved! You have taken two abstract planes, found their hidden orientations, combined them using the power of the cross product, and anchored the result to a specific point in space.

This is the beauty of 3D geometry—it turns abstract equations into a tangible, structural reality. Keep practicing, keep visualizing, and remember: every complex problem is just a series of elegant, logical steps waiting for you to connect them.