The system of equations 3x + 2y + z=6, 3x + 4y + 3z = 14, 6x + 10y + 8z = a, has infinite number of solutions, if a =

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Solution

The correct option is D 36
It will have infinite soln if $Δ = Δ_{1} = Δ_{2} = Δ_{3} = 0$
$Δ = ∣ ∣ ∣ ∣ \begin{matrix} 3 & 2 & 1 3 & 4 & 3 6 & 10 & 8 \end{matrix} ∣ ∣ ∣ ∣ = 3 \times 2 ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 1 & 2 & 3 2 & 5 & 8 \end{matrix} ∣ ∣ ∣ ∣ = 6 [1 + (- 2) + 1] = 0$
$Δ_{1} = ∣ ∣ ∣ ∣ \begin{matrix} 6 & 2 & 2 14 & 4 & 3 a & 10 & 8 \end{matrix} ∣ ∣ ∣ ∣ = 6 (2) + 2 (3 a - 112) + 140 - 4 a$
$∵ Δ_{1} = 0$
$12 + 6 a = 224 + 140 - 4 a = 0$
$\Rightarrow 2 a = 224 - 152$
$\Rightarrow a = 112 - 76$
$\Rightarrow a = 36$

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