If `4x+3y+12z=26, x, y, z, in R`, then minimum possible value of `x^(2)+y^(2)+z^(2)` is __________

Answer

To find the minimum possible value of \( x^2 + y^2 + z^2 \) given the equation \( 4x + 3y + 12z = 26 \), we can use the method of Lagrange multipliers or geometric interpretation. Here, we will use the geometric interpretation involving the distance from a point to a plane.

### Step-by-Step Solution:

1. **Identify the Plane Equation**:
The equation \( 4x + 3y + 12z = 26 \) represents a plane in three-dimensional space.

2. **Distance from the Origin to the Plane**:
We want to find the minimum distance from the origin (0, 0, 0) to this plane. The formula for the distance \( D \) from a point \( (x_1, y_1, z_1) \) to the plane \( Ax + By + Cz = D \) is given by:
\[ D = \frac{|Ax_1 + By_1 + Cz_1 - D|}{\sqrt{A^2 + B^2 + C^2}} \]
Here, \( A = 4 \), \( B = 3 \), \( C = 12 \), and \( D = 26 \). We will substitute \( (x_1, y_1, z_1) = (0, 0, 0) \).

3. **Calculate the Distance**:
Substituting into the formula:
\[ D = \frac{|4(0) + 3(0) + 12(0) - 26|}{\sqrt{4^2 + 3^2 + 12^2}} = \frac{|0 - 26|}{\sqrt{16 + 9 + 144}} = \frac{26}{\sqrt{169}} = \frac{26}{13} = 2 \]

4. **Finding the Minimum Value of \( x^2 + y^2 + z^2 \)**:
The minimum value of \( x^2 + y^2 + z^2 \) occurs at the point on the plane that is closest to the origin, which corresponds to the square of the distance we just calculated:
\[ x^2 + y^2 + z^2 = D^2 = 2^2 = 4 \]

5. **Conclusion**:
Therefore, the minimum possible value of \( x^2 + y^2 + z^2 \) is \( \boxed{4} \).