Animated Solution for Mathematics - Matrices and Determinants: Let the system of linear equations $-x+2y-9z=7$, $-x+3y+7z=9$, $-2x+y+5z=8$, $-3x+y+13z=\lambda$ has a unique solution $x=\alpha ,y=\beta ,z=\gamma$. Then the distance of the point $(\alpha ,\beta ,\gamma )$ from the plane $2x–2y+z=\lambda$ is

Welcome, future engineer. Today, we are not just solving a system of equations; we are peeling back the layers of a 3D geometric puzzle.

When you look at a system like this, don't see it as a chore of algebra. See it as the intersection of planes in space. Each equation represents a flat surface, and the 'unique solution' is the singular point where these surfaces meet in perfect harmony.

The Mystery of the Fourth Equation

We are presented with four equations:

(i) $-x+2y-9z=7$

(ii) $-x+3y+7z=9$

(iii) $-2x+y+5z=8$

(iv) $-3x+y+13z=\lambda$

I can hear the hesitation. 'Four equations, three variables?' you ask. 'Is this a trick?'

It is not a trick; it is a test of your confidence. If a system has a unique solution, the first three equations are all we need to define the point $P(\alpha ,\beta ,\gamma )$.

The fourth equation is merely a gatekeeper, waiting for us to find the coordinates so it can reveal the value of $\lambda$. We shall ignore it for now and focus our energy on the first three.

The Elimination Dance

To find the point, we must systematically reduce our variables. Let us target $x$ first. Look at equations (i) and (ii). They both feature a $-x$ term. This is a gift!

Subtracting (i) from (ii) gives us:

Which simplifies beautifully to:

Now, let us turn our attention to equation (iii). It has a $-2x$. To eliminate $x$ here, we need to match it. Multiply equation (i) by $2$ to get $-2x+4y-18z=14$.

Now, subtract equation (iii) from this result:

This leaves us with:

The Unveiling

We have reduced our complex system to a simple $2\times 2$ system: $y+16z=2$ and $3y-23z=6$. From (v), we know $y=2-16z$.

Let us substitute this into (vi):

Notice the elegance here? The constants cancel out! We are left with $-71z=0$, which forces $z=0$.

With $z=0$, finding $y$ is trivial: $y=2-16(0)=2$. Finally, plugging $y=2$ and $z=0$ into equation (i), we find $-x+2(2)-9(0)=7$, which leads to $-x+4=7$, or $x=-3$.

Our unique point $P$ is $(-3,2,0)$.

The Bridge to Geometry

Now, we return to the fourth equation. Since $P(-3,2,0)$ is the solution to the entire system, it must satisfy equation (iv).

Substituting our values:

We have unlocked the parameter. Our plane is $2x-2y+z=11$, or $2x-2y+z-11=0$.

The Final Leap

We are at the finish line. We need the perpendicular distance from $P(-3,2,0)$ to the plane $2x-2y+z-11=0$. Recall the formula:

Plugging in our values:

The distance is 7. Take a moment to appreciate the journey. We navigated through algebraic elimination, solved for the unknown parameter, and applied geometric principles to find the distance. You didn't just solve a problem; you mastered a process.