Which of the following systems has only a trivial solution?

The correct answer is Option (3) → \( x+y+z=0, \)
\( 4x+2y+z=0, \)
\( 9x+3y+z=0 \)

A homogeneous system (all constant terms zero) always has the trivial solution x = y = z = 0. It has only the trivial solution when the determinant of its coefficients is non-zero.

Option (1) \( 4x-2y+9z+2=0, \)
\( 3x+4y+z-5=0, \)
\( x-3y+2z-8=0 \) — Incorrect. It has constant terms, so it is not homogeneous and x = y = z = 0 does not even satisfy it.

Option (2) \( x+y-z=0, \)
\( x-2y+z=0, \)
\( 3x+6y-5z=0 \) — Incorrect. The determinant is 0, so there are non-trivial solutions:

    \( \displaystyle \begin{vmatrix}1&1&-1\\1&-2&1\\3&6&-5\end{vmatrix}=1(10-6)-1(-5-3)-1(6+6) \)

    \( \displaystyle =4+8-12=0 \)

Option (3) \( x+y+z=0, \)
\( 4x+2y+z=0, \)
\( 9x+3y+z=0 \) — Correct. The determinant is non-zero:

    \( \displaystyle \begin{vmatrix}1&1&1\\4&2&1\\9&3&1\end{vmatrix}=1(2-3)-1(4-9)+1(12-18) \)

    \( \displaystyle =-1+5-6=-2 \)

Option (4) \( 2x-3y-z=0, \)
\( x+3y-2z=0, \)
\( x-3y=0 \) — Incorrect. The determinant is 0, so there are non-trivial solutions:

    \( \displaystyle \begin{vmatrix}2&-3&-1\\1&3&-2\\1&-3&0\end{vmatrix}=2(0-6)+3(0+2)-1(-3-3) \)

    \( \displaystyle =-12+6+6=0 \)