If `vecx and vecy` are two non zero, non - collinear vectors satisfying
`((a-3)alpha^(2)+(b-4)alpha+(c-1))vecx +[(a-3)beta^(2)+(b-4)beta+(c-1)]vecy+[(a-3)gamma^(2)+(b-4)gamma+(c-1)](vecx xx vecy)=0` (where `alpha, beta,gamma` are three distinct numbers), then the value of `(a^(2)+b^(2)+c^(2))/(4)` is equal to

Answer

To solve the problem, we need to analyze the given vector equation step by step.

### Step 1: Understand the given equation
The equation provided is:
\[ ((a-3)\alpha^2 + (b-4)\alpha + (c-1)\vec{x} + [(a-3)\beta^2 + (b-4)\beta + (c-1)]\vec{y} + [(a-3)\gamma^2 + (b-4)\gamma + (c-1)](\vec{x} \times \vec{y}) = 0 \]
This equation implies that the vectors \(\vec{x}\), \(\vec{y}\), and \(\vec{x} \times \vec{y}\) must be coplanar.

### Step 2: Set up conditions for coplanarity
For the vectors to be coplanar, each coefficient of the vectors must equal zero. This gives us three equations:
1. \((a-3)\alpha^2 + (b-4)\alpha + (c-1) = 0\)
2. \((a-3)\beta^2 + (b-4)\beta + (c-1) = 0\)
3. \((a-3)\gamma^2 + (b-4)\gamma + (c-1) = 0\)

### Step 3: Recognize the quadratic nature
The equations are quadratic in nature, with \(\alpha\), \(\beta\), and \(\gamma\) being the roots.

### Step 4: Establish the identity condition
For a quadratic equation to have three distinct roots, it must be an identity, meaning all coefficients must be zero:
1. Coefficient of \(\alpha^2\): \(a - 3 = 0\)
2. Coefficient of \(\alpha\): \(b - 4 = 0\)
3. Constant term: \(c - 1 = 0\)

### Step 5: Solve for \(a\), \(b\), and \(c\)
From the above conditions, we can solve:
- \(a - 3 = 0 \Rightarrow a = 3\)
- \(b - 4 = 0 \Rightarrow b = 4\)
- \(c - 1 = 0 \Rightarrow c = 1\)

### Step 6: Calculate \(a^2 + b^2 + c^2\)
Now we can find \(a^2 + b^2 + c^2\):
\[ a^2 + b^2 + c^2 = 3^2 + 4^2 + 1^2 = 9 + 16 + 1 = 26 \]

### Step 7: Find the final result
We need to find \(\frac{a^2 + b^2 + c^2}{4}\):
\[ \frac{26}{4} = 6.5 \]

### Final Answer
Thus, the value of \(\frac{a^2 + b^2 + c^2}{4}\) is:
\[ \boxed{6.5} \]