If the system of linear equations
$x_{1} + 2 x_{2} + 3 x_{3} = 6;$
$x_{1} + 3 x_{2} + 5 x_{3} = 9;$
$2 x_{1} + 5 x_{2} + a x_{3} = b$
is consistent and has infinite number of solutions, then

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a = 8, b can be any real number

b = 15, a can be any real number

$a \in R - \{8\}$ and $b \in R - \{15\}$

a = 8, b = 15

answer is D.

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Detailed Solution

Infinite number of solutions $D = 0, D_{1} = 0, D_{2} = 0, D_{3} = 0$

$| \begin{array}{lll} 1 & 2 & 3 \\ 1 & 3 & 5 \\ 2 & 5 & a \end{array} | = 0$

$1 (3 a - 25) - 2 (a - 10) + 3 (- 1) = 0$

$a - 25 + 20 - 3 = 0$

$a - 8 = 0 \Rightarrow a = 8$

$D_{1} = 0$

$| \begin{array}{lll} 6 & 2 & 3 \\ 9 & 3 & 5 \\ b & 5 & 8 \end{array} | = 0$

$6 (- 1) - 2 (72 - 5 b) + 3 (45 - 3 b) = 0$

$- 6 - 144 - 10 b + 135 - 9 b = 0$

$b = 15$

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