Key Concepts and Formulas

This problem combines concepts from Matrix Algebra (specifically matrix multiplication), Systems of Linear Equations, and 3D Geometry (the condition for a point to lie on a plane).

1. Matrix Multiplication: If we have a row matrix $A = \begin{pmatrix} a & b & c \end{pmatrix}$ and a $3 \times 3$ matrix $M = \begin{pmatrix} m_{11} & m_{12} & m_{13} \\ m_{21} & m_{22} & m_{23} \\ m_{31} & m_{32} & m_{33} \end{pmatrix}$, their product $AM$ will be a row matrix of dimension $1 \times 3$. The elements are calculated as follows: $AM = \begin{pmatrix} (a \cdot m_{11} + b \cdot m_{21} + c \cdot m_{31}) & (a \cdot m_{12} + b \cdot m_{22} + c \cdot m_{32}) & (a \cdot m_{13} + b \cdot m_{23} + c \cdot m_{33}) \end{pmatrix}$
2. Point on a Plane: A point $P(x_0, y_0, z_0)$ lies on a plane $Ax+By+Cz=D$ if and only if its coordinates satisfy the plane's equation, i.e., $Ax_0+By_0+Cz_0=D$.
3. Solving System of Linear Equations: We use methods like substitution or elimination to find the values of unknown variables from a set of linear equations.

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Step-by-Step Solution with Explanations

Step 1: Formulate the equation from the plane condition.

We are given that the point $P(\alpha, \beta, \gamma)$ lies on the plane $2x+4y+3z=5$.

• Explanation: For a point to lie on a plane, its coordinates must satisfy the equation of the plane. We substitute $\alpha$ for $x$, $\beta$ for $y$, and $\gamma$ for $z$ into the plane equation.

Substituting the coordinates of $P$ into the plane equation, we get: $2\alpha + 4\beta + 3\gamma = 5 \quad \text{.........(1)}$

Step 2: Derive equations from the matrix multiplication.

We are given the matrix equation: \left( {\matrix{ \alpha & \beta & \gamma \cr } } \right)\left( {\matrix{ 2 & {10} & 8 \cr 9 & 3 & 8 \cr 8 & 4 & 8 \cr } } \right) = \left( {\matrix{ 0 & 0 & 0 \cr } } \right)

• Explanation: We perform the matrix multiplication on the left side. The product of a $1 \times 3$ row matrix and a $3 \times 3$ square matrix will result in a $1 \times 3$ row matrix. Each element of this resultant matrix is then equated to the corresponding element of the zero matrix on the right side.

Let's perform the matrix multiplication: The first element of the resultant matrix is $(\alpha \cdot 2 + \beta \cdot 9 + \gamma \cdot 8)$. The second element is $(\alpha \cdot 10 + \beta \cdot 3 + \gamma \cdot 4)$. The third element is $(\alpha \cdot 8 + \beta \cdot 8 + \gamma \cdot 8)$.

Equating these to the elements of the zero matrix \left( {\matrix{ 0 & 0 & 0 \cr } } \right), we get three linear equations:

1. $2\alpha + 9\beta + 8\gamma = 0 \quad \text{.........(2)}$
2. $10\alpha + 3\beta + 4\gamma = 0 \quad \text{.........(3)}$
3. $8\alpha + 8\beta + 8\gamma = 0 \quad \text{.........(4)}$

Step 3: Solve the system of linear equations to find $\alpha, \beta, \gamma$.

We now have a system of four linear equations (1), (2), (3), and (4) involving $\alpha, \beta, \gamma$. Our goal is to find unique values for these variables.

• Explanation: We look for opportunities to simplify equations or eliminate variables. Equation (4) is particularly simple as all coefficients are 8.

From equation (4), we can divide by 8: $\frac{8\alpha + 8\beta + 8\gamma}{8} = \frac{0}{8}$ $\alpha + \beta + \gamma = 0 \quad \text{.........(4')}$ This simplified equation is very useful as it allows us to express one variable in terms of the others. For instance, $\gamma = -\alpha - \beta$.

Now, let's use equations (2) and (4') to find a relationship between $\alpha$ and $\beta$. Substitute $\gamma = -\alpha - \beta$ into equation (2): $2\alpha + 9\beta + 8(-\alpha - \beta) = 0$ $2\alpha + 9\beta - 8\alpha - 8\beta = 0$ $-6\alpha + \beta = 0$ $\beta = 6\alpha \quad \text{.........(5)}$

• Explanation: By substituting $\gamma$ from (4') into (2), we eliminated $\gamma$, allowing us to express $\beta$ solely in terms of $\alpha$. This is a common strategy in solving systems of equations.

Now, substitute $\beta = 6\alpha$ into equation (4'): $\alpha + (6\alpha) + \gamma = 0$ $7\alpha + \gamma = 0$ $\gamma = -7\alpha \quad \text{.........(6)}$

• Explanation: We have now expressed both $\beta$ and $\gamma$ in terms of $\alpha$. These relations define the general solution for the homogeneous system from the matrix equation. This means any point $(\alpha, \beta, \gamma)$ satisfying the matrix equation must be of the form $(k, 6k, -7k)$ for some scalar $k$.

Finally, substitute the expressions for $\beta$ (from (5)) and $\gamma$ (from (6)) into equation (1) (the plane equation) to find the specific value of $\alpha$: $2\alpha + 4\beta + 3\gamma = 5$ $2\alpha + 4(6\alpha) + 3(-7\alpha) = 5$ $2\alpha + 24\alpha - 21\alpha = 5$ Combine the terms involving $\alpha$: $(2 + 24 - 21)\alpha = 5$ $5\alpha = 5$ $\alpha = 1$

• Explanation: This step allowed us to find the specific value of $\alpha$. Now we can find $\beta$ and $\gamma$.

Using $\alpha=1$ in equations (5) and (6): $\beta = 6\alpha = 6(1) = 6$ $\gamma = -7\alpha = -7(1) = -7$ So, the point $P$ is $(1, 6, -7)$.

Step 4: Calculate the required expression.

We need to find the value of $6\alpha+9\beta+7\gamma$.

• Explanation: Now that we have the values of $\alpha, \beta, \gamma$, we simply substitute them into the given expression and calculate the result.

Substitute $\alpha=1$, $\beta=6$, and $\gamma=-7$: $6\alpha + 9\beta + 7\gamma = 6(1) + 9(6) + 7(-7)$ $= 6 + 54 - 49$ $= 60 - 49$ $= 11$

Thus, $6\alpha+9\beta+7\gamma = 11$.

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Common Mistakes & Tips

1. Matrix Multiplication Accuracy: Be very careful with matrix multiplication. A common mistake is getting the order of multiplication or the sum of products incorrect. Always remember (row $\times$ column).
2. Simplify Equations: Always look for opportunities to simplify equations, like dividing by a common factor (as we did with equation (4)). This makes subsequent calculations easier and reduces the chance of errors.
3. Systematic Elimination/Substitution: When solving systems of linear equations, choose a systematic approach. Expressing variables in terms of one another (e.g., $\beta$ and $\gamma$ in terms of $\alpha$) often simplifies the process.

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Summary

This problem effectively tests your understanding of fundamental concepts in linear algebra and 3D geometry. The key steps involved were: translating the geometric condition (point on a plane) into an algebraic equation, translating the matrix equation into a system of linear equations, solving the resulting system of linear equations using techniques like substitution and elimination to find the coordinates $(\alpha, \beta, \gamma)$, and finally, evaluating the required expression. A systematic and careful approach to solving simultaneous equations is crucial for success in such problems. Based on the given problem statement, the calculated value is 11.

The final answer is $\boxed{\text{11}}$, which corresponds to option (B).