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

From equation (1): $4x^2 + 4y^2 + 4z^2 = a$

From equation (2): $4x - 4y - 4z = 3 + a$

Since both equations equal expressions involving 'a', subtracting will eliminate 'a' completely.

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

$4x^2 + 4y^2 + 4z^2 - 4x + 4y + 4z = -3$

---

$4x^2 - 4x + 4y^2 + 4y + 4z^2 + 4z = -3$

The big idea: This expression looks like it wants to be written as perfect squares! This is a common technique when we see $ax^2 + bx$ terms.

---

To complete the square for $4x^2 - 4x$, we want $(2x - ?)^2$

For $4x^2 - 4x$: We need $(2x - 1)^2 = 4x^2 - 4x + 1$

For $4y^2 + 4y$: We need $(2y + 1)^2 = 4y^2 + 4y + 1$

For $4z^2 + 4z$: We need $(2z + 1)^2 = 4z^2 + 4z + 1$

Each completion requires adding 1. So we add $1 + 1 + 1 = 3$ to both sides:

$(4x^2 - 4x + 1) + (4y^2 + 4y + 1) + (4z^2 + 4z + 1) = -3 + 3$

$(2x-1)^2 + (2y+1)^2 + (2z+1)^2 = 0$

---

The sum of squares of real numbers equals zero only when each square is zero.

Since $(2x-1)^2 \geq 0$, $(2y+1)^2 \geq 0$, and $(2z+1)^2 \geq 0$ for all real numbers, and their sum is 0, each must equal 0:

$(2x-1)^2 = 0 \rightarrow 2x - 1 = 0 \rightarrow x = \frac{1}{2}$

$(2y+1)^2 = 0 \rightarrow 2y + 1 = 0 \rightarrow y = -\frac{1}{2}$

$(2z+1)^2 = 0 \rightarrow 2z + 1 = 0 \rightarrow z = -\frac{1}{2}$

---

We substitute our values into equation (1):

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

$a = 4\left(\left(\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2\right)$

$a = 4\left(\frac{1}{4} + \frac{1}{4} + \frac{1}{4}\right) = 4 \times \frac{3}{4} = 3$

Therefore, $a = 3$