Key Concepts and Formulas

• Homogeneous System of Linear Equations: A system of linear equations is homogeneous if all constant terms are zero, typically written as $AX = 0$.
  • It always has the trivial solution $x=0, y=0, z=0$.
  • It has non-trivial solutions (solutions where at least one variable is non-zero) if and only if the determinant of the coefficient matrix $A$ is zero, i.e., $\det(A) = 0$. In this case, there are infinitely many solutions, often representing a line passing through the origin.
  • If $\det(A) \neq 0$, the only solution is the trivial solution $(0,0,0)$.
• Method of Cross-Multiplication (for two equations in three variables): For two linear equations $A_1x + B_1y + C_1z = 0$ and $A_2x + B_2y + C_2z = 0$, the general solution can be found using the ratio: $\frac{x}{B_1C_2 - B_2C_1} = \frac{y}{C_1A_2 - C_2A_1} = \frac{z}{A_1B_2 - A_2B_1} = k$ where $k$ is a proportionality constant (parameter).
• Point on a Plane: A point $(x_0, y_0, z_0)$ lies on a plane $Ax + By + Cz = D$ if substituting its coordinates into the plane equation satisfies the equation, i.e., $Ax_0 + By_0 + Cz_0 = D$.

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Step-by-Step Solution

Step 1: Identify the System of Equations and Form the Coefficient Matrix. The given system of linear equations is:

1. $x + 8y + 7z = 0$
2. $9x + 2y + 3z = 0$
3. $x + y + z = 0$

Since all constant terms on the right-hand side are zero, this is a homogeneous system of linear equations. The coefficient matrix $A$ for this system is: $A = \begin{pmatrix} 1 & 8 & 7 \\ 9 & 2 & 3 \\ 1 & 1 & 1 \end{pmatrix}$

Step 2: Determine the Nature of Solutions by Calculating the Determinant of the Coefficient Matrix. To find out if there are non-trivial solutions, we calculate the determinant of matrix $A$: $\det(A) = 1 \begin{vmatrix} 2 & 3 \\ 1 & 1 \end{vmatrix} - 8 \begin{vmatrix} 9 & 3 \\ 1 & 1 \end{vmatrix} + 7 \begin{vmatrix} 9 & 2 \\ 1 & 1 \end{vmatrix}$ Expanding the $2 \times 2$ determinants: $\det(A) = 1((2)(1) - (3)(1)) - 8((9)(1) - (3)(1)) + 7((9)(1) - (2)(1))$ $\det(A) = 1(2 - 3) - 8(9 - 3) + 7(9 - 2)$ $\det(A) = 1(-1) - 8(6) + 7(7)$ $\det(A) = -1 - 48 + 49$ $\det(A) = 0$ Explanation: Since $\det(A) = 0$, the system of homogeneous linear equations has infinitely many non-trivial solutions. This means the three planes represented by the equations intersect along a common line passing through the origin. Our goal is to find the equation of this line.

Step 3: Find the General Parametric Solution for $(x,y,z)$. Because the determinant is zero, the equations are linearly dependent, meaning any two linearly independent equations can be used to find the general solution. Let's choose equations (1) and (3) for simplicity:

1. $x + 8y + 7z = 0$
2. $x + y + z = 0$

We can use the method of cross-multiplication with coefficients: For equation (1): $A_1=1, B_1=8, C_1=7$ For equation (3): $A_2=1, B_2=1, C_2=1$

Applying the cross-multiplication formula: $\frac{x}{B_1C_2 - B_2C_1} = \frac{y}{C_1A_2 - C_2A_1} = \frac{z}{A_1B_2 - A_2B_1}$ Calculate the denominators:

• For $x$: $(8)(1) - (1)(7) = 8 - 7 = 1$
• For $y$: $(7)(1) - (1)(1) = 7 - 1 = 6$
• For $z$: $(1)(1) - (1)(8) = 1 - 8 = -7$

So, the general solution for $(x,y,z)$ is: $\frac{x}{1} = \frac{y}{6} = \frac{z}{-7}$ Let this common ratio be $k$. Then, we can express $x, y, z$ in terms of $k$: $x = k$ $y = 6k$ $z = -7k$ Thus, the solution $(a,b,c)$ to the system of equations is $(k, 6k, -7k)$ for some real number $k$.

Verification: To ensure correctness, let's substitute this general solution into the second equation (which we didn't use to derive the general solution): $9x + 2y + 3z = 9(k) + 2(6k) + 3(-7k) = 9k + 12k - 21k = 21k - 21k = 0$. The general solution satisfies all three equations, confirming its validity.

Step 4: Use the Additional Condition that the Point $(a,b,c)$ Lies on the Plane $x + 2y + z = 6$. We are given that $(a,b,c)$ is a solution to the system, so $(a,b,c) = (k, 6k, -7k)$. We are also given that this point lies on the plane $x + 2y + z = 6$. Substitute the parametric coordinates into the plane equation: $(k) + 2(6k) + (-7k) = 6$ $k + 12k - 7k = 6$ Combine the terms: $(1 + 12 - 7)k = 6$ $6k = 6$ $k = 1$

Step 5: Determine the Specific Values of $a, b, c$ and Calculate $2a+b+c$. Now that we have $k=1$, we can find the specific values of $a, b, c$: $a = k = 1$ $b = 6k = 6(1) = 6$ $c = -7k = -7(1) = -7$ So, the specific solution is $(a,b,c) = (1, 6, -7)$.

Finally, we need to calculate the value of the expression $2a + b + c$: $2a + b + c = 2(1) + 6 + (-7)$ $2a + b + c = 2 + 6 - 7$ $2a + b + c = 8 - 7$ $2a + b + c = 1$

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Common Mistakes & Tips

• Assuming Trivial Solution: A common mistake is to assume $x=y=z=0$ is the only solution for a homogeneous system. Always check the determinant of the coefficient matrix. If $\det(A)=0$, non-trivial solutions exist and must be found.
• Calculation Errors in Determinant or Cross-Multiplication: These are prone to sign errors or arithmetic mistakes. Double-check calculations, especially when dealing with multiple terms.
• Forgetting to Verify: After finding the general parametric solution from two equations, always substitute it into the third (unused) equation to ensure it satisfies all conditions. This catches potential errors in the cross-multiplication or choice of equations.

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Summary

This problem required us to first analyze a system of homogeneous linear equations. By calculating the determinant of the coefficient matrix, we found it to be zero, indicating infinitely many non-trivial solutions. We then used the method of cross-multiplication on two of the equations to find the general parametric form of these solutions as $(k, 6k, -7k)$. Finally, we utilized the given condition that this solution point lies on the plane $x + 2y + z = 6$ to determine the specific value of the parameter $k=1$. This allowed us to find the unique solution $(a,b,c) = (1, 6, -7)$ and subsequently calculate the desired expression $2a+b+c$, which resulted in 1.

The final answer is $\boxed{1}$, which corresponds to option (C).