1. Key Concepts and Formulas

• Direction Vector of a Line: The direction vector of a line passing through two points $Q_1(x_1, y_1, z_1)$ and $Q_2(x_2, y_2, z_2)$ is given by $\vec{d} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$.
• Normal Vector of a Plane: If a plane is perpendicular to a line, then the direction vector of that line serves as the normal vector ($\vec{n}$) to the plane. The components of the normal vector $(a, b, c)$ are the coefficients of $x, y, z$ in the plane's equation.
• Equation of a Plane: The equation of a plane passing through a point $(x_0, y_0, z_0)$ with a normal vector $a\hat{i} + b\hat{j} + c\hat{k}$ is given by: $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$
• Point on a Plane: If a point $(\lambda, y_p, z_p)$ lies on a plane, its coordinates must satisfy the equation of the plane.

2. Step-by-Step Solution

Step 1: Determine the Normal Vector of the Plane

• What we are doing: We are given that the plane is perpendicular to the line joining points $Q_1(-2, -21, 29)$ and $Q_2(-1, -16, 23)$.
• Why we are doing it: According to our key concept, the direction vector of this line will be the normal vector to the plane. This normal vector is essential for writing the plane's equation.
• Working: Let the direction vector of the line be $\vec{n}$. $\vec{n} = \vec{Q_2} - \vec{Q_1}$ $\vec{n} = (-1 - (-2))\hat{i} + (-16 - (-21))\hat{j} + (23 - 29)\hat{k}$ $\vec{n} = (1)\hat{i} + (5)\hat{j} + (-6)\hat{k}$ So, the normal vector to the plane is $\vec{n} = \hat{i} + 5\hat{j} - 6\hat{k}$. Thus, the coefficients for the plane equation are $a=1$, $b=5$, $c=-6$.

Step 2: Formulate the Equation of the Plane

• What we are doing: We now have the normal vector $\vec{n} = \hat{i} + 5\hat{j} - 6\hat{k}$ and a point $P_1(4, -2, 2)$ that the plane passes through.
• Why we are doing it: We use the standard formula for the equation of a plane to find its algebraic representation.
• Working: Using the point $P_1(x_0, y_0, z_0) = (4, -2, 2)$ and the normal vector components $(a, b, c) = (1, 5, -6)$: $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$ $1(x - 4) + 5(y - (-2)) + (-6)(z - 2) = 0$ $1(x - 4) + 5(y + 2) - 6(z - 2) = 0$ Expand the equation: $x - 4 + 5y + 10 - 6z + 12 = 0$ Combine the constant terms: $x + 5y - 6z + (10 + 12 - 4) = 0$ $x + 5y - 6z + 18 = 0$ This is the equation of the plane.

Step 3: Find the value of $\lambda$

• What we are doing: We are given that the point $(\lambda, 2, 1)$ lies on the plane.
• Why we are doing it: Since the point lies on the plane, its coordinates must satisfy the plane's equation. Substituting these coordinates will allow us to solve for $\lambda$.
• Working: Substitute $x=\lambda$, $y=2$, and $z=1$ into the plane equation $x + 5y - 6z + 18 = 0$: $\lambda + 5(2) - 6(1) + 18 = 0$ $\lambda + 10 - 6 + 18 = 0$ $\lambda + 4 + 18 = 0$ $\lambda + 22 = 0$ $\lambda = -22$

Step 4: Calculate the value of the given expression

• What we are doing: We need to evaluate the expression ${\left( {{\lambda \over {11}}} \right)^2} - {{4\lambda } \over {11}} - 4$ using the value of $\lambda$ we just found.
• Why we are doing it: This is the final requirement of the problem statement.
• Working: Substitute $\lambda = -22$ into the expression: ${\left( {{-22 \over {11}}} \right)^2} - {{4(-22)} \over {11}} - 4$ First, simplify the terms with $\lambda/11$: ${(-2)^2} - {4(-2)} - 4$ Calculate the squares and products: $4 - (-8) - 4$ $4 + 8 - 4$ $12 - 4$ $8$

3. Common Mistakes & Tips

• Sign Errors: Be extremely careful with negative signs when calculating direction vectors and substituting coordinates into the plane equation. A common mistake is $y - (-2)$ becoming $y-2$.
• Order of Operations: Ensure correct order of operations (parentheses, exponents, multiplication/division, addition/subtraction) when evaluating the final expression.
• Vector vs. Scalar: Remember that the normal vector provides the direction ratios $(a, b, c)$, which are scalar coefficients in the plane equation.

4. Summary

We first determined the normal vector of the plane by finding the direction vector of the line to which the plane is perpendicular. Using this normal vector and a given point on the plane, we formulated the equation of the plane. Then, by substituting the coordinates of the point $(\lambda, 2, 1)$ into the plane equation, we solved for $\lambda$. Finally, we substituted the value of $\lambda$ into the given algebraic expression to obtain the required numerical value.

5. Final Answer

The final answer is $\boxed{8}$.