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 and extract the necessary information step by step.

### Step 1: Understand the given equation
We have the equation:
\[ ((a-3)\alpha^2 + (b-4)\alpha + (c-1)\vec{x} + [(a-3)\beta^2 + (b-4)\beta + (c-1)]\vec{y} + [(a-3)\gamma^2 + (b-4)\gamma + (c-1)](\vec{x} \times \vec{y}) = 0 \]
This equation holds for three distinct values of \(\alpha\), \(\beta\), and \(\gamma\).

### Step 2: Set up individual equations
Since \(\vec{x}\) and \(\vec{y}\) are non-zero and non-collinear, the only way for the entire expression to equal zero is if each coefficient of the vectors \(\vec{x}\), \(\vec{y}\), and \(\vec{x} \times \vec{y}\) is equal to zero. Thus, we can set up the following equations:
1. \(a - 3\alpha^2 + b - 4\alpha + c - 1 = 0\) (for \(\vec{x}\))
2. \(a - 3\beta^2 + b - 4\beta + c - 1 = 0\) (for \(\vec{y}\))
3. \(a - 3\gamma^2 + b - 4\gamma + c - 1 = 0\) (for \(\vec{x} \times \vec{y}\))

### Step 3: Recognize the quadratic nature
The equations represent a quadratic polynomial in terms of \(x\):
\[ (a-3)x^2 + (b-4)x + (c-1) = 0 \]
with roots \(\alpha\), \(\beta\), and \(\gamma\). For a quadratic polynomial to have three roots, it must be identically zero, which implies that all coefficients must be zero.

### Step 4: Solve for \(a\), \(b\), and \(c\)
From the coefficients, we can set up the following equations:
1. \(a - 3 = 0 \implies a = 3\)
2. \(b - 4 = 0 \implies b = 4\)
3. \(c - 1 = 0 \implies c = 1\)

### Step 5: Calculate \(a^2 + b^2 + c^2\)
Now, we can calculate:
\[ a^2 + b^2 + c^2 = 3^2 + 4^2 + 1^2 = 9 + 16 + 1 = 26 \]

### Step 6: Find \(\frac{a^2 + b^2 + c^2}{4}\)
Finally, we compute:
\[ \frac{a^2 + b^2 + c^2}{4} = \frac{26}{4} = 6.5 \]

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

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