Su of values of P for which equation x + y + z = 1,x + 2y 4z = P and x + 4y + 10z = $p^{2}$ has a solution is

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Solution

The correct option is C 3
Given equation,
$x + y + z = 1 x + 2 y + 4 z = p x + 4 y + 10 z = p^{2}$

$Δ = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 1 & 2 & 4 1 & 4 & 10 \end{matrix} ∣ ∣ ∣ ∣$

Applying $R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3}$

$= ∣ ∣ ∣ ∣ \begin{matrix} 0 & - 1 & - 3 0 & - 2 & - 6 1 & 4 & 10 \end{matrix} ∣ ∣ ∣ ∣$

Since $Δ = 0$ Solution is not unique solution

The system will have infinite solution if $Δ_{1} = 0, Δ_{1} = 0, Δ_{3} = 0$

$Δ_{1} = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 p & 2 & 4 p^{2} & 4 & 10 \end{matrix} ∣ ∣ ∣ ∣$

Applying $C_{3} \to C_{3} - C_{2}$

$⟹ 1 (12 - 8) - 1 (6 p - 2 p^{2}) = 0$

$⟹ p^{2} - 3 p + 2 = 0$

$⟹ p = 1, p = 2$

hence sum of values of $p = 1 + 2 = 3$

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