Step 1: Analyze the given condition.
\(\vec{a} \times (\vec{c} - \vec{b}) = 0 \implies \vec{a} || (\vec{c} - \vec{b})\).
Thus, \(\vec{c} - \vec{b} = k \vec{a}\), where \(k\) is a scalar.
Step 2: Solve for \(\vec{c}\).
\[ \vec{c} = \vec{b} + k\vec{a} = (a + 6k)\hat{i} + (11 + 9k)\hat{j} + (-2 + 12k)\hat{k}. \] Step 3: Use the given conditions.
1. \(\vec{a} \cdot \vec{c} = -12\): \[ 6(a + 6k) + 9(11 + 9k) + 12(-2 + 12k) = -12. \] Simplify to solve for \(a\) and \(k\).
2. Substituting \(\vec{c}\) into \(\vec{c} \cdot (\hat{i} - 2\hat{j} + \hat{k}) = 5\): \[ (a + 6k) - 2(11 + 9k) + (-2 + 12k) = 5. \] Solving these, we find \(\vec{c} = 23\hat{i} + 2\hat{j} - 14\hat{k}\).
Step 4: Compute \(\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k})\).
\[ \vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 23 + 2 - 14 = 11. \] Final Answer: \(\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 11\).
Let the vectors \(\mathbf{u}_1 = \hat{i} + \hat{j} + a\hat{k}, \mathbf{u}_2 = \hat{i} + b\hat{j} + \hat{k}\), and \(\mathbf{u}_3 = c\hat{i} + \hat{j} + \hat{k}\) be coplanar. If the vectors \(\mathbf{v}_1 = (a + b)\hat{i} + c\hat{j} + c\hat{k}, \mathbf{v}_2 = a\hat{i} + (b + c)\hat{j} + a\hat{k}, \mathbf{v}_3 = b\hat{i} + b\hat{j} + (c + a)\hat{k}\) are also coplanar, then \(6(a + b + c)\) is equal to:
If the points with position vectors \(a\hat{i} +10\hat{j} +13\hat{k}, 6\hat{i} +11\hat{k} +11\hat{k},\frac{9}{2}\hat{i}+B\hat{j}−8\hat{k}\) are collinear, then (19α-6β)2 is equal to
Let S be the set of all (λ, μ) for which the vectors $ λ {i}ˆ-jˆ+kˆ, iˆ +2jˆ+µkˆ and 3iˆ -4jˆ +5kˆ, where λ-μ = 5, are coplanar, then $$ \sum_{(λ, μ) εs}80(λ^2, μ^2) $ is equal to