Given that `vecA,vecB,vecC` form triangle such that `vecA=vecB+vecC`. Find a,b,c,d such that area of the triangle is `5sqrt(6)` where `vecA=aveci+bveci+cveck. vecB=dveci+3vecj+4veck and vecC=3veci+vecj-2veck`.

Explanation

To solve the problem step by step, we need to find the values of \(a\), \(b\), \(c\), and \(d\) such that the area of the triangle formed by the vectors \(\vec{A}\), \(\vec{B}\), and \(\vec{C}\) is \(5\sqrt{6}\). Given the relationships between the vectors, we can follow these steps:

### Step 1: Write down the relationships between the vectors
We know that:
\[ \vec{A} = \vec{B} + \vec{C} \]
Given:
\[ \vec{A} = a\vec{i} + b\vec{j} + c\vec{k} \]
\[ \vec{B} = d\vec{i} + 3\vec{j} + 4\vec{k} \]
\[ \vec{C} = 3\vec{i} + \vec{j} - 2\vec{k} \]

### Step 2: Substitute \(\vec{B}\) and \(\vec{C}\) into the equation for \(\vec{A}\)
Substituting the expressions for \(\vec{B}\) and \(\vec{C}\) into the equation for \(\vec{A}\):
\[ a\vec{i} + b\vec{j} + c\vec{k} = (d + 3)\vec{i} + (3 + 1)\vec{j} + (4 - 2)\vec{k} \]
This simplifies to:
\[ a = d + 3, \quad b = 4, \quad c = 2 \]

### Step 3: Calculate the area of the triangle using the cross product
The area \(A\) of the triangle formed by vectors \(\vec{B}\) and \(\vec{C}\) is given by:
\[ A = \frac{1}{2} |\vec{B} \times \vec{C}| \]
We need to find \(|\vec{B} \times \vec{C}|\) and set it equal to \(10\sqrt{6}\) (since \(5\sqrt{6} = \frac{1}{2} |\vec{B} \times \vec{C}|\)).

### Step 4: Compute the cross product \(\vec{B} \times \vec{C}\)
Using the determinant formula for the cross product:
\[ \vec{B} \times \vec{C} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ d & 3 & 4 \\ 3 & 1 & -2 \end{vmatrix} \]
Calculating this determinant:
\[ \vec{B} \times \vec{C} = \vec{i}(3 \cdot (-2) - 4 \cdot 1) - \vec{j}(d \cdot (-2) - 4 \cdot 3) + \vec{k}(d \cdot 1 - 3 \cdot 3) \]
This simplifies to:
\[ \vec{B} \times \vec{C} = \vec{i}(-6 - 4) - \vec{j}(-2d - 12) + \vec{k}(d - 9) \]
\[ = -10\vec{i} + (2d + 12)\vec{j} + (d - 9)\vec{k} \]

### Step 5: Find the magnitude of the cross product
The magnitude is given by:
\[ |\vec{B} \times \vec{C}| = \sqrt{(-10)^2 + (2d + 12)^2 + (d - 9)^2} \]
Setting this equal to \(10\sqrt{6}\):
\[ \sqrt{100 + (2d + 12)^2 + (d - 9)^2} = 10\sqrt{6} \]
Squaring both sides:
\[ 100 + (2d + 12)^2 + (d - 9)^2 = 600 \]

### Step 6: Expand and simplify the equation
Expanding:
\[ (2d + 12)^2 = 4d^2 + 48d + 144 \]
\[ (d - 9)^2 = d^2 - 18d + 81 \]
Combining these:
\[ 100 + 4d^2 + 48d + 144 + d^2 - 18d + 81 = 600 \]
This simplifies to:
\[ 5d^2 + 30d + 325 = 600 \]
\[ 5d^2 + 30d - 275 = 0 \]
Dividing through by 5:
\[ d^2 + 6d - 55 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula:
\[ d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-55)}}{2 \cdot 1} \]
\[ = \frac{-6 \pm \sqrt{36 + 220}}{2} = \frac{-6 \pm \sqrt{256}}{2} = \frac{-6 \pm 16}{2} \]
Calculating the two possible values:
\[ d = \frac{10}{2} = 5 \quad \text{or} \quad d = \frac{-22}{2} = -11 \]

### Step 8: Find corresponding values of \(a\), \(b\), and \(c\)
Using \(d = 5\):
\[ a = 5 + 3 = 8, \quad b = 4, \quad c = 2 \]
Using \(d = -11\):
\[ a = -11 + 3 = -8, \quad b = 4, \quad c = 2 \]

### Final Answer
Thus, the values are:
1. For \(d = 5\): \(a = 8\), \(b = 4\), \(c = 2\)
2. For \(d = -11\): \(a = -8\), \(b = 4\), \(c = 2\)