Key Concepts and Formulas

• Coplanarity of Vectors: Three vectors $\overrightarrow{p}$, $\overrightarrow{q}$, and $\overrightarrow{r}$ are coplanar if their scalar triple product (STP) is zero. The STP is given by the determinant of their components: $[\overrightarrow{p} \ \overrightarrow{q} \ \overrightarrow{r}] = \overrightarrow{p} \cdot (\overrightarrow{q} \times \overrightarrow{r}) = 0$.
• Dot Product: For vectors $\overrightarrow{A} = A_x\widehat{i} + A_y\widehat{j} + A_z\widehat{k}$ and $\overrightarrow{B} = B_x\widehat{i} + B_y\widehat{j} + B_z\widehat{k}$, $\overrightarrow{A} \cdot \overrightarrow{B} = A_x B_x + A_y B_y + A_z B_z$.
• Cross Product: For vectors $\overrightarrow{A}$ and $\overrightarrow{B}$, $\overrightarrow{A} \times \overrightarrow{B} = \left| {\begin{array}{ccc} \widehat i & \widehat j & \widehat k \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{array}} \right|$.
• Magnitude Squared of a Vector: For a vector $\overrightarrow{V} = V_x\widehat{i} + V_y\widehat{j} + V_z\widehat{k}$, $|\overrightarrow{V}|^2 = V_x^2 + V_y^2 + V_z^2$.

Step-by-Step Solution

Step 1: Determine the value of 'a' using the coplanarity condition. The vectors $\overrightarrow p$, $\overrightarrow q$, and $\overrightarrow r$ are given by: $\overrightarrow p = \left( {a + 1} \right)\widehat i + a\widehat j + a\widehat k$ $\overrightarrow q = a\widehat i + \left( {a + 1} \right)\widehat j + a\widehat k$ $\overrightarrow r = a\widehat i + a\widehat j + \left( {a + 1} \right)\widehat k$ For these vectors to be coplanar, their scalar triple product must be zero: $[\overrightarrow p \ \overrightarrow q \ \overrightarrow r] = \left| {\begin{array}{ccc} a+1 & a & a \\ a & a+1 & a \\ a & a & a+1 \end{array}} \right| = 0$ To simplify the determinant, we apply the column operation $C_1 \to C_1 + C_2 + C_3$: $\left| {\begin{array}{ccc} (a+1)+a+a & a & a \\ a+(a+1)+a & a+1 & a \\ a+a+(a+1) & a & a+1 \end{array}} \right| = \left| {\begin{array}{ccc} 3a+1 & a & a \\ 3a+1 & a+1 & a \\ 3a+1 & a & a+1 \end{array}} \right| = 0$ Factor out $(3a+1)$ from the first column: $(3a+1) \left| {\begin{array}{ccc} 1 & a & a \\ 1 & a+1 & a \\ 1 & a & a+1 \end{array}} \right| = 0$ Now, apply row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$: $(3a+1) \left| {\begin{array}{ccc} 1 & a & a \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}} \right| = 0$ Expanding the determinant along the first column: $(3a+1) [1 \cdot (1 \cdot 1 - 0 \cdot 0)] = 0$ $(3a+1)(1) = 0 \implies 3a+1 = 0 \implies a = -\frac{1}{3}$

Step 2: Calculate $(\overrightarrow p \cdot \overrightarrow q)^2$. First, compute the dot product $\overrightarrow p \cdot \overrightarrow q$: $\overrightarrow p \cdot \overrightarrow q = ((a+1)\widehat i + a\widehat j + a\widehat k) \cdot (a\widehat i + (a+1)\widehat j + a\widehat k)$ $\overrightarrow p \cdot \overrightarrow q = (a+1)(a) + (a)(a+1) + (a)(a)$ $\overrightarrow p \cdot \overrightarrow q = a^2+a + a^2+a + a^2 = 3a^2 + 2a$ Substitute $a = -\frac{1}{3}$: $\overrightarrow p \cdot \overrightarrow q = 3\left(-\frac{1}{3}\right)^2 + 2\left(-\frac{1}{3}\right) = 3\left(\frac{1}{9}\right) - \frac{2}{3} = \frac{1}{3} - \frac{2}{3} = -\frac{1}{3}$ Now, square the result: $(\overrightarrow p \cdot \overrightarrow q)^2 = \left(-\frac{1}{3}\right)^2 = \frac{1}{9}$

Step 3: Calculate $|\overrightarrow r \times \overrightarrow q|^2$. First, compute the cross product $\overrightarrow r \times \overrightarrow q$: $\overrightarrow r \times \overrightarrow q = \left| {\begin{array}{ccc} \widehat i & \widehat j & \widehat k \\ a & a & a+1 \\ a & a+1 & a \end{array}} \right|$ Expanding the determinant: $\overrightarrow r \times \overrightarrow q = \widehat i [a(a) - (a+1)(a+1)] - \widehat j [a(a) - (a+1)(a)] + \widehat k [a(a+1) - a(a)]$ $\overrightarrow r \times \overrightarrow q = \widehat i [a^2 - (a^2+2a+1)] - \widehat j [a^2 - (a^2+a)] + \widehat k [a^2+a - a^2]$ $\overrightarrow r \times \overrightarrow q = \widehat i (-2a-1) - \widehat j (-a) + \widehat k (a)$ $\overrightarrow r \times \overrightarrow q = -(2a+1)\widehat i + a\widehat j + a\widehat k$ Now, calculate the magnitude squared: $|\overrightarrow r \times \overrightarrow q|^2 = (-(2a+1))^2 + (a)^2 + (a)^2$ $|\overrightarrow r \times \overrightarrow q|^2 = (2a+1)^2 + 2a^2 = 4a^2+4a+1 + 2a^2 = 6a^2+4a+1$ Substitute $a = -\frac{1}{3}$: $|\overrightarrow r \times \overrightarrow q|^2 = 6\left(-\frac{1}{3}\right)^2 + 4\left(-\frac{1}{3}\right) + 1$ $|\overrightarrow r \times \overrightarrow q|^2 = 6\left(\frac{1}{9}\right) - \frac{4}{3} + 1 = \frac{2}{3} - \frac{4}{3} + 1 = -\frac{2}{3} + 1 = \frac{1}{3}$

Step 4: Substitute the calculated values into the given equation and solve for $\lambda$. The given equation is: $3{\left( {\overrightarrow p .\overrightarrow q } \right)^2} - \lambda \left| {\overrightarrow r \times \overrightarrow q } \right|^2 = 0$ Substitute the values $(\overrightarrow p \cdot \overrightarrow q)^2 = \frac{1}{9}$ and $|\overrightarrow r \times \overrightarrow q|^2 = \frac{1}{3}$: $3\left(\frac{1}{9}\right) - \lambda \left(\frac{1}{3}\right) = 0$ $\frac{1}{3} - \frac{\lambda}{3} = 0$ Multiply by 3: $1 - \lambda = 0$ $\lambda = 1$

Common Mistakes & Tips

• Sign Errors in Cross Product: Be very careful with the signs when expanding the determinant for the cross product, especially the $\widehat j$ component.
• Forgetting to Square: The problem requires squaring the dot product and the magnitude of the cross product. Ensure these operations are performed correctly.
• Algebraic Simplification: Using determinant properties and row/column operations can significantly reduce the complexity of calculations.

Summary The problem requires using the coplanarity condition to find the value of $a$. Once $a$ is determined, we calculate the dot product $(\overrightarrow p \cdot \overrightarrow q)^2$ and the squared magnitude of the cross product $|\overrightarrow r \times \overrightarrow q|^2$. Finally, these values are substituted into the given equation to solve for $\lambda$. The coplanarity condition leads to $a = -1/3$. Calculating the required vector operations with this value of $a$ and substituting them into the equation $3{\left( {\overrightarrow p .\overrightarrow q } \right)^2} - \lambda \left| {\overrightarrow r \times \overrightarrow q } \right|^2 = 0$ yields $\lambda = 1$.

The final answer is $\boxed{1}$.