Let $$ x, y $$, and $$ z $$ be real numbers satisfying:

$$ 4(x^2 + y^2 + z^2) = a $$
$$ 4(x - y - z) = 3 + a $$

Then, $$ a $$ equals:

Question

Let $$ x, y $$, and $$ z $$ be real numbers satisfying:

$$ 4(x^2 + y^2 + z^2) = a $$  
$$ 4(x - y - z) = 3 + a $$

Then, $$ a $$ equals:

Accepted Answer

The problem involves solving for the value of $$a$$ under the given conditions involving real numbers $$x$$, $$y$$, and $$z$$. Here's the step-by-step solution:

**Given Equations:**

- $$4(x^2 + y^2 + z^2) = a$$
   - $$4(x - y - z) = 3 + a$$

Simplify these equations:
   - $$x^2 + y^2 + z^2 = \frac{a}{4}$$
   - $$x - y - z = \frac{3 + a}{4}$$

**Apply the Identity:**

The identity $$x^2 + y^2 + z^2 \geq xy + yz + zx$$ holds for real numbers. Expanding:
   $$
   (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)
   $$
   From the identity:
   $$
   (x + y + z)^2 \leq 3(x^2 + y^2 + z^2)
   $$
   Substituting values:
   - From $$x - y - z = \frac{3 + a}{4}$$, let $$y_1 = -y$$ and $$z_1 = -z$$. Then:
     $$
     x + y_1 + z_1 = \frac{3 + a}{4}
     $$
   - Substitute into the inequality:
     $$
     \left(\frac{3 + a}{4}\right)^2 \leq 3\left(\frac{a}{4}\right)
     $$

**Simplify the Inequality:**

Expand both sides:
   $$
   \frac{(3 + a)^2}{16} \leq \frac{3a}{4}
   $$
   Multiply through by 16 to eliminate the denominator:
   $$
   (3 + a)^2 \leq 12a
   $$
   Expand:
   $$
   9 + a^2 + 6a \leq 12a
   $$
   Rearrange terms:
   $$
   a^2 - 6a + 9 \leq 0
   $$
   Factorize:
   $$
   (a - 3)^2 \leq 0
   $$

**Solve for $$a$$:**

The inequality $$ (a - 3)^2 \leq 0 $$ is satisfied only when:
   $$
   a = 3
   $$

### Final Answer:

The value of $$a$$ is **3**.  
Correct option: **A**

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