Let x, y, and z be real numbers satisfying

4(x2 + y2 + z2) = a,

4(x – y – z) = 3 + a

The a equals

The correct option is 3.

We are given:

4(x² + y² + z²) = a ...(1)

4(x − y − z) = 3 + a ...(2)

From Equation (2),

a = 4(x − y − z) − 3

Substituting this value of a into Equation (1),

4(x² + y² + z²) = 4(x − y − z) − 3

Expanding and bringing all terms to the left-hand side,

4x² + 4y² + 4z² − 4x + 4y + 4z + 3 = 0

Grouping the terms,

(4x² − 4x + 1) + (4y² + 4y + 1) + (4z² + 4z + 1) = 0

Therefore,

(2x − 1)² + (2y + 1)² + (2z + 1)² = 0

Since x, y, and z are real numbers, each squared term is non-negative. Hence, their sum can be zero only when each term is zero.

2x − 1 = 0 ⇒ x = 1/2

2y + 1 = 0 ⇒ y = −1/2

2z + 1 = 0 ⇒ z = −1/2

Using Equation (1),

a = 4(x² + y² + z²)

Substituting the values of x, y, and z,

a = 4[(1/2)² + (−1/2)² + (−1/2)²]

= 4(1/4 + 1/4 + 1/4)

= 4 × 3/4

= 3

Thus, a = 3.

Hence, the correct option is 3.