To find the value of a that makes the system of equations have no solution, we need to ensure that the determinant of the coefficient matrix is zero. This will indicate that the equations are linearly dependent, leading to no unique solutions. Given the system of equations: x+2y+3z‌=6 x+3y+5z‌=9 2x+5y+az‌=12 we consider the coefficient matrix: [
1
2
3
1
3
5
2
5
a
] Calculate the determinant of this matrix: D=|
1
2
3
1
3
5
2
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a
| The determinant is calculated as follows: D‌=1⋅(3a−25)−2⋅(a−10)+3⋅(5−6) ‌=3a−25−2a+20−3 ‌=3a−2a−25+20−3 ‌=a−8 Setting the determinant D=0 gives: a−8=0, which simplifies to: a=8‌. ‌ Therefore, the system of equations has no solution when a=8.