Key Concepts and Formulas

• Scalar Multiplication of a Vector: If $\overrightarrow v = v_x \widehat i + v_y \widehat j + v_z \widehat k$ and $k$ is a scalar, then $k\overrightarrow v = (kv_x)\widehat i + (kv_y)\widehat j + (kv_z)\widehat k$.
• Equality of Vectors: Two vectors are equal if and only if their corresponding components are equal.
• Dot Product of Perpendicular Vectors: Two non-zero vectors $\overrightarrow A$ and $\overrightarrow B$ are perpendicular if and only if their dot product is zero, i.e., $\overrightarrow A \cdot \overrightarrow B = 0$. For $\overrightarrow A = A_x \widehat i + A_y \widehat j + A_z \widehat k$ and $\overrightarrow B = B_x \widehat i + B_y \widehat j + B_z \widehat k$, the dot product is $\overrightarrow A \cdot \overrightarrow B = A_x B_x + A_y B_y + A_z B_z$.

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Step-by-Step Solution

We are given three vectors: $\overrightarrow a = 2\widehat i + {\lambda _1}\widehat j + 3\widehat k$ $\overrightarrow b = 4\widehat i + \left( {3 - {\lambda _2}} \right)\widehat j + 6\widehat k$ $\overrightarrow c = 3\widehat i + 6\widehat j + \left( {{\lambda _3} - 1} \right)\widehat k$

Step 1: Apply the condition $\overrightarrow b = 2\overrightarrow a$

The condition $\overrightarrow b = 2\overrightarrow a$ implies that vector $\overrightarrow b$ is obtained by multiplying each component of vector $\overrightarrow a$ by the scalar 2. First, let's compute $2\overrightarrow a$: $2\overrightarrow a = 2(2\widehat i + {\lambda _1}\widehat j + 3\widehat k) = (2 \times 2)\widehat i + (2 \times {\lambda _1})\widehat j + (2 \times 3)\widehat k$ $2\overrightarrow a = 4\widehat i + 2{\lambda _1}\widehat j + 6\widehat k$ Now, we equate the components of $\overrightarrow b$ with the components of $2\overrightarrow a$: $4\widehat i + \left( {3 - {\lambda _2}} \right)\widehat j + 6\widehat k = 4\widehat i + 2{\lambda _1}\widehat j + 6\widehat k$ Comparing the $\widehat i$ components: $4 = 4$. Comparing the $\widehat j$ components: $3 - {\lambda _2} = 2{\lambda _1}$. Rearranging this equation, we get: $2{\lambda _1} + {\lambda _2} = 3 \quad \text{(Equation 1)}$ Comparing the $\widehat k$ components: $6 = 6$.

Step 2: Apply the condition $\overrightarrow a \perp \overrightarrow c$

The condition that $\overrightarrow a$ is perpendicular to $\overrightarrow c$ means their dot product is zero. The dot product $\overrightarrow a \cdot \overrightarrow c$ is calculated as: $\overrightarrow a \cdot \overrightarrow c = (2)(3) + ({\lambda _1})(6) + (3)({\lambda _3} - 1)$ $\overrightarrow a \cdot \overrightarrow c = 6 + 6{\lambda _1} + 3{\lambda _3} - 3$ $\overrightarrow a \cdot \overrightarrow c = 6{\lambda _1} + 3{\lambda _3} + 3$ Since $\overrightarrow a \perp \overrightarrow c$, we set the dot product to zero: $6{\lambda _1} + 3{\lambda _3} + 3 = 0$ Dividing the entire equation by 3 for simplification: $2{\lambda _1} + {\lambda _3} + 1 = 0$ Rearranging this equation, we get: $2{\lambda _1} + {\lambda _3} = -1 \quad \text{(Equation 2)}$

Step 3: Test the given options against the derived equations

We have two equations relating $\lambda_1, \lambda_2,$ and $\lambda_3$:

1. $2{\lambda _1} + {\lambda _2} = 3$
2. $2{\lambda _1} + {\lambda _3} = -1$

We need to find an option $(\lambda_1, \lambda_2, \lambda_3)$ that satisfies both these equations.

(A) $(1, 5, 1)$ For Equation 1: $2(1) + 5 = 2 + 5 = 7 \neq 3$. This option is incorrect.

(B) $(1, 3, 1)$ For Equation 1: $2(1) + 3 = 2 + 3 = 5 \neq 3$. This option is incorrect.

(C) $\left( - {1 \over 2}, 4, 0 \right)$ For Equation 1: $2\left( -{1 \over 2} \right) + 4 = -1 + 4 = 3$. This matches Equation 1. For Equation 2: $2\left( -{1 \over 2} \right) + 0 = -1 + 0 = -1$. This matches Equation 2. Since both equations are satisfied, this option is a possible value.

(D) $\left( {{1 \over 2}, 4, - 2} \right)$ For Equation 1: $2\left( {1 \over 2} \right) + 4 = 1 + 4 = 5 \neq 3$. This option is incorrect.

The only option that satisfies both conditions is (C).

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Common Mistakes & Tips

• Component-wise Errors: Ensure that when equating vectors or applying scalar multiplication, all corresponding components are correctly handled.
• Dot Product Calculation: Double-check the arithmetic when calculating the dot product and setting it to zero for perpendicular vectors.
• System of Equations: Recognize that with more variables than independent equations, there might be infinite solutions. The question asks for a possible value, so testing options is a valid strategy.

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Summary

The problem requires applying the definitions of scalar multiplication of vectors, vector equality, and the dot product for perpendicular vectors. By using the condition $\overrightarrow b = 2\overrightarrow a$, we derived a linear relationship between $\lambda_1$ and $\lambda_2$. By using the condition $\overrightarrow a \perp \overrightarrow c$, we derived a linear relationship between $\lambda_1$ and $\lambda_3$. Finally, we tested the given options against these derived equations to find the one that satisfies both conditions. Option (C) was found to satisfy both derived equations.

The final answer is $\boxed{\left( { - {1 \over 2},4,0} \right)}$, which corresponds to option (C).