Consider the system of linear equations \[ x+y+z=1, \]\[ 2x+y+3z=6, \]\[ 3x+2y+kz=k+1. \]If the above system has no solution, then the value of \(k\) equals _______ (answer in integer).
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Eliminate \(x\) and \(y\) to get an equation of the form \((k-4)z=k-6\); the system has no solution when this reduces to \(0=\text{nonzero}\), that is at \(k=4\).
Step 1: Set up the coefficient picture. The system is \[ x+y+z=1,\quad 2x+y+3z=6,\quad 3x+2y+kz=k+1 \] Notice that the coefficients of $x$ and $y$ in the third equation, namely $3$ and $2$, equal the sum of the coefficients in the first two equations: $1+2=3$ for $x$ and $1+1=2$ for $y$. This hints that adding equation (i) and equation (ii) will connect directly to equation (iii).
Step 2: Add the first two equations. \[ (x+y+z)+(2x+y+3z)=1+6 \] \[ 3x+2y+4z=7 \]
Step 3: Compare with the third equation. The third equation is $3x+2y+kz=k+1$. Since the $x$ and $y$ coefficients in this sum, $3$ and $2$, exactly match the third equation's coefficients, subtract the third equation from the sum found in Step 2: \[ (3x+2y+4z)-(3x+2y+kz)=7-(k+1) \] \[ (4-k)z=6-k \]
Step 4: Read off when this breaks down. This is the same relation as before, just written with signs flipped: $(4-k)z=6-k$ is the same equation as $z(k-4)=k-6$. If $k\ne4$, this fixes a unique $z$, and the first two original equations then fix $x$ and $y$ uniquely, so a solution exists and is unique.
Step 5: Check the special value $k=4$. Putting $k=4$ into $(4-k)z=6-k$ gives \[ (4-4)z=6-4 \] \[ 0=2 \] which is impossible for any $z$. So at $k=4$ the three planes represented by the equations do not share a common point, meaning the system has no solution. This also matches the geometric fact that the first two planes intersect in a line, and at $k=4$ the third plane becomes parallel to that line without containing it.
Step 6: State the answer. No solution occurs exactly at \[ \boxed{4} \]
