Key Concepts and Formulas

• Coplanarity of Vectors: Three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ are coplanar if their scalar triple product is zero, i.e., $[\vec{a} \ \vec{b} \ \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c}) = 0$.
• Scalar Triple Product (STP) as a Determinant: If $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$, $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$, and $\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$, then their STP is given by the determinant: $[\vec{a} \ \vec{b} \ \vec{c}] = \left|\begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array}\right|$
• Properties of Determinants:
  • Adding a multiple of one row (or column) to another row (or column) does not change the determinant's value.
  • Taking a common factor from a row (or column) out of the determinant.

Step-by-Step Solution

Step 1: Applying the coplanarity condition to $\vec{u}_1, \vec{u}_2, \vec{u}_3$. We are given that $\vec{u}_{1}=\hat{i}+\hat{j}+a \hat{k}$, $\vec{u}_{2}=\hat{i}+b \hat{j}+\hat{k}$, and $\vec{u}_{3}=c \hat{i}+\hat{j}+\hat{k}$ are coplanar. This means their scalar triple product is zero. We can express this as a determinant of their components: $[\vec{u}_1 \ \vec{u}_2 \ \vec{u}_3] = \left|\begin{array}{lll} 1 & 1 & a \\ 1 & b & 1 \\ c & 1 & 1 \end{array}\right|=0$ Expanding this determinant along the first row: $1 \cdot (b \cdot 1 - 1 \cdot 1) - 1 \cdot (1 \cdot 1 - c \cdot 1) + a \cdot (1 \cdot 1 - b \cdot c) = 0$ $1(b-1) - 1(1-c) + a(1-bc) = 0$ $b-1-1+c+a-abc = 0$ $a+b+c-2-abc = 0$ $a+b+c-2 = abc \quad \ldots(1)$

Step 2: Applying the coplanarity condition to $\vec{v}_1, \vec{v}_2, \vec{v}_3$. We are given that $\vec{v}_{1}=(a+b) \hat{i}+c \hat{j}+c \hat{k}$, $\vec{v}_{2}=a \hat{i}+(b+c) \hat{j}+a \hat{k}$, and $\vec{v}_{3}=b \hat{i}+b \hat{j}+(c+a) \hat{k}$ are also coplanar. Therefore, their scalar triple product is zero: $[\vec{v}_1 \ \vec{v}_2 \ \vec{v}_3] = \left|\begin{array}{ccc} a+b & c & c \\ a & b+c & a \\ b & b & c+a \end{array}\right|=0$ To simplify this determinant, we use row operations. Let $D$ be the determinant. Apply $R_1 \to R_1 + R_2 + R_3$: $D = \left|\begin{array}{ccc} (a+b)+a+b & c+(b+c)+b & c+a+(c+a) \\ a & b+c & a \\ b & b & c+a \end{array}\right| = \left|\begin{array}{ccc} 2(a+b) & 2(b+c) & 2(c+a) \\ a & b+c & a \\ b & b & c+a \end{array}\right|$ Factor out 2 from the first row: $D = 2 \left|\begin{array}{ccc} a+b & b+c & c+a \\ a & b+c & a \\ b & b & c+a \end{array}\right|$ Now, apply $R_1 \to R_1 - R_2$: $D = 2 \left|\begin{array}{ccc} (a+b)-a & (b+c)-(b+c) & (c+a)-a \\ a & b+c & a \\ b & b & c+a \end{array}\right| = 2 \left|\begin{array}{ccc} b & 0 & c \\ a & b+c & a \\ b & b & c+a \end{array}\right|$ Expand the determinant along the first row: $D = 2 \left[ b \cdot \left|\begin{array}{cc} b+c & a \\ b & c+a \end{array}\right| - 0 \cdot \left|\begin{array}{cc} a & a \\ b & c+a \end{array}\right| + c \cdot \left|\begin{array}{cc} a & b+c \\ b & b \end{array}\right| \right]$ $D = 2 \left[ b((b+c)(c+a) - ab) + c(ab - b(b+c)) \right]$ $D = 2 \left[ b(bc+ba+c^2+ca - ab) + c(ab - b^2 - bc) \right]$ $D = 2 \left[ b(bc+c^2+ca) + abc - b^2c - bc^2 \right]$ $D = 2 \left[ b^2c+bc^2+abc + abc - b^2c - bc^2 \right]$ $D = 2 \left[ 2abc \right]$ $D = 4abc$ Since the vectors are coplanar, $D=0$, so $4abc = 0$. This implies $abc = 0 \quad \ldots(2)$.

Step 3: Combining the results to find the value of $6(a+b+c)$. We have two equations:

1. $a+b+c-2 = abc$
2. $abc = 0$ Substitute equation (2) into equation (1): $a+b+c-2 = 0$ $a+b+c = 2$ The problem asks for the value of $6(a+b+c)$. $6(a+b+c) = 6(2) = 12$

Common Mistakes & Tips

• Determinant Expansion Errors: Be very careful with the signs and arithmetic when expanding determinants. Using row/column operations to simplify the determinant before expansion is highly recommended.
• Algebraic Simplification: Ensure meticulous algebraic manipulation, especially when dealing with multiple variables. Grouping terms and cancelling them out systematically can prevent errors.
• Understanding Coplanarity: The core concept is that coplanar vectors have a zero scalar triple product. This is the bridge between vector algebra and determinant properties.

Summary

The problem requires applying the condition of coplanarity of vectors, which translates to the scalar triple product being zero. We first set up a determinant for the given vectors $\vec{u}_1, \vec{u}_2, \vec{u}_3$ and obtained an algebraic relation. Then, we did the same for vectors $\vec{v}_1, \vec{v}_2, \vec{v}_3$, using row operations to simplify the determinant, which led to a simpler relation. By combining these two algebraic relations, we found the value of $a+b+c$ and subsequently calculated the required expression $6(a+b+c)$.

The final answer is $\boxed{\text{12}}$.