Key Concept: Conditions for Infinitely Many Solutions in a System of Linear Equations

For a system of linear equations of the form: $a_1x + b_1y + c_1z = d_1$ $a_2x + b_2y + c_2z = d_2$ $a_3x + b_3y + c_3z = d_3$ to have infinitely many solutions, the determinant of the coefficient matrix, denoted by $\Delta$, must be zero. Additionally, the determinants $\Delta_x$, $\Delta_y$, and $\Delta_z$ (formed by replacing the respective column of coefficients with the constant terms) must also be zero. That is, $\Delta = 0, \quad \Delta_x = 0, \quad \Delta_y = 0, \quad \Delta_z = 0$

If $\Delta = 0$ but at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero, the system has no solution.

Step-by-Step Derivation

1. Set up the System of Equations: The given system of linear equations is: $8x + y + 4z = -2 \quad \text{(Equation 1)}$ $x + y + z = 0 \quad \text{(Equation 2)}$ $\lambda x - 3y + 0z = \mu \quad \text{(Equation 3)}$

2. Calculate the Determinant of the Coefficient Matrix ($\Delta$): We first form the coefficient matrix and calculate its determinant. $\Delta = \left| {\begin{matrix} 8 & 1 & 4 \\ 1 & 1 & 1 \\ \lambda & { - 3} & 0 \end{matrix}} \right|$ To expand this determinant, we can use cofactor expansion along the third row (or any row/column). Let's expand along the third row for variety, or the given solution uses first row, let's stick to that. Expanding along the first row: $\Delta = 8 \left| {\begin{matrix} 1 & 1 \\ -3 & 0 \end{matrix}} \right| - 1 \left| {\begin{matrix} 1 & 1 \\ \lambda & 0 \end{matrix}} \right| + 4 \left| {\begin{matrix} 1 & 1 \\ \lambda & -3 \end{matrix}} \right|$ $\Delta = 8((1)(0) - (1)(-3)) - 1((1)(0) - (1)(\lambda)) + 4((1)(-3) - (1)(\lambda))$ $\Delta = 8(0 + 3) - 1(0 - \lambda) + 4(-3 - \lambda)$ $\Delta = 8(3) - 1(-\lambda) + 4(-3 - \lambda)$ $\Delta = 24 + \lambda - 12 - 4\lambda$ $\Delta = 12 - 3\lambda$

3. Apply the Condition for Infinitely Many Solutions to find $\lambda$: For the system to have infinitely many solutions, $\Delta$ must be zero. $12 - 3\lambda = 0$ $3\lambda = 12$ $\lambda = 4$

4. Calculate $\Delta_x$ and Apply the Condition to find $\mu$: Next, we calculate $\Delta_x$ by replacing the first column (coefficients of $x$) with the constant terms of the equations. $\Delta_x = \left| {\begin{matrix} -2 & 1 & 4 \\ 0 & 1 & 1 \\ \mu & -3 & 0 \end{matrix}} \right|$ Expanding this determinant along the first column: $\Delta_x = -2 \left| {\begin{matrix} 1 & 1 \\ -3 & 0 \end{matrix}} \right| - 0 \left| {\begin{matrix} 1 & 4 \\ -3 & 0 \end{matrix}} \right| + \mu \left| {\begin{matrix} 1 & 4 \\ 1 & 1 \end{matrix}} \right|$ $\Delta_x = -2((1)(0) - (1)(-3)) - 0 + \mu((1)(1) - (4)(1))$ $\Delta_x = -2(0 + 3) + \mu(1 - 4)$ $\Delta_x = -2(3) + \mu(-3)$ $\Delta_x = -6 - 3\mu$ For infinitely many solutions, $\Delta_x$ must also be zero. $-6 - 3\mu = 0$ $3\mu = -6$ $\mu = -2$

• Important Tip: While we only calculated $\Delta_x$ here, it's crucial to understand that for infinitely many solutions, all determinants $\Delta_x, \Delta_y, \Delta_z$ must be zero. If we were to calculate $\Delta_y$ and $\Delta_z$ with $\lambda=4$ and $\mu=-2$, we would indeed find them to be zero, confirming the consistency of the system.

5. Determine the Point and the Plane: From our calculations, we found $\lambda = 4$ and $\mu = -2$. The given point is $P = \left( {\lambda ,\mu , - {1 \over 2}} \right)$, which becomes $P = \left( {4, -2, - {1 \over 2}} \right)$. The given plane is $8x + y + 4z + 2 = 0$.

6. Calculate the Distance of the Point from the Plane:

The formula for the perpendicular distance of a point $(x_1, y_1, z_1)$ from a plane $Ax + By + Cz + D = 0$ is: $\text{Distance} = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$ Here, $(x_1, y_1, z_1) = \left( {4, -2, - {1 \over 2}} \right)$ and the plane coefficients are $A=8, B=1, C=4, D=2$.

Substitute these values into the formula: $\text{Distance} = \frac{|8(4) + 1(-2) + 4\left(-{1 \over 2}\right) + 2|}{\sqrt{8^2 + 1^2 + 4^2}}$ $\text{Distance} = \frac{|32 - 2 - 2 + 2|}{\sqrt{64 + 1 + 16}}$ $\text{Distance} = \frac{|30|}{\sqrt{81}}$ $\text{Distance} = \frac{30}{9}$ $\text{Distance} = \frac{10}{3}$

• Common Mistake: Forgetting the absolute value in the numerator or making arithmetic errors during determinant expansion or distance calculation.

Summary and Key Takeaway

To find the values of parameters for which a system of linear equations has infinitely many solutions, we must ensure that the determinant of the coefficient matrix ($\Delta$) and all the auxiliary determinants ($\Delta_x, \Delta_y, \Delta_z$) are zero. Once these parameters are found, we can then use standard geometric formulas, such as the distance from a point to a plane, to complete the problem.

The distance of the point $\left( {4, -2, - {1 \over 2}} \right)$ from the plane $8x + y + 4z + 2 = 0$ is $\frac{10}{3}$ units.

The final answer is $\boxed{\text{D}}$.