Animated Solution for Mathematics - Matrices and Determinants: If the system of linear equations $8x+y+4z=-2$, $x+y+z=0$, $\lambda x-3y=\mu$ has infinitely many solutions, then the distance of the point $(\lambda ,\mu ,-\frac{1}{2})$ from the plane $8x+y+4z+2=0$ is :

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a system of equations; we are exploring the architecture of 3D space.

Imagine standing in a room where three massive, flat planes intersect. Usually, they meet at a single point—a corner of a room.

But here, the problem tells us something fascinating: the system has infinitely many solutions. This means our three planes do not meet at a single point, but instead, they slice through each other to form a common line of intersection.

When we face a system of linear equations with parameters like $\lambda$ and $\mu$, we need a powerful lens. Enter Cramer's Rule.

For a system to have infinitely many solutions, the 'volume' of the parallelepiped formed by the coefficients must collapse to zero. Mathematically, this means the main determinant $D$ and the auxiliary determinants $D_1$, $D_2$, and $D_3$ must all vanish.

We are looking for the condition where the system becomes dependent, where one equation is essentially a linear combination of the others.

Let us set up our determinant $D$ using the coefficients of $x$, $y$, and $z$:

We expand along the first row:

Simplifying this, we find $24+\lambda -12-4\lambda =0$, which leads to $12-3\lambda =0$. Thus, $\lambda =4$.

With $\lambda$ secured, we repeat the process for $D_1$ by replacing the first column with the constants $-2$, $0$, and $\mu$:

Expanding this yields:

This simplifies to $-6+\mu -4\mu =0$, giving us $\mu =-2$. We have successfully cracked the code.

Now that we have our parameters, we identify our point $P$ as $(4,-2,-\frac{1}{2})$. The final challenge is to find the distance of this point from the plane $8x+y+4z+2=0$.

Notice that this plane is actually the first equation from our original system. We use the classic distance formula:

Substituting our coordinates, we get:

The numerator simplifies to $|32-2-2+2|=30$, and the denominator is $\sqrt{64+1+16}=\sqrt{81}=9$.

Thus, the final distance is:

Look at that result: $\frac{10}{3}$. It is not just a number; it is the culmination of understanding how planes interact in space.

You navigated the determinants, handled the parameters, and applied the distance formula with precision. Keep this geometric intuition alive—it is the secret weapon of every great mathematician.