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My approach is as follow

$D = \left| {\begin{array}{*{20}{c}} {2\left( {a + b} \right)}&{3\left( {b + c} \right)}&{4\left( {a + c} \right)}\\ {2\left( {b + c} \right)}&{3\left( {a + c} \right)}&{4\left( {a + b} \right)}\\ {2\left( {a + c} \right)}&{3\left( {a + b} \right)}&{4\left( {b + c} \right)} \end{array}} \right|$

$\alpha = a + b;\beta = b + c,\gamma = a + c$

$D = \left| {\begin{array}{*{20}{c}} {2\alpha }&{3\beta }&{4\gamma }\\ {2\beta }&{3\gamma }&{4\alpha }\\ {2\gamma }&{3\alpha }&{4\beta } \end{array}} \right| \Rightarrow D = 2 \times 3 \times 4\left| {\begin{array}{*{20}{c}} \alpha &\beta &\gamma \\ \beta &\gamma &\alpha \\ \gamma &\alpha &\beta \end{array}} \right| \Rightarrow D = 24\left| {\begin{array}{*{20}{c}} \alpha &\beta &\gamma \\ \beta &\gamma &\alpha \\ \gamma &\alpha &\beta \end{array}} \right|$

$ \Rightarrow D = 24\left[ {\alpha \left( {\beta \gamma - {\alpha ^2}} \right) - \beta \left( {{\beta ^2} - \alpha \gamma } \right) + \gamma \left( {\alpha \beta - {\gamma ^2}} \right)} \right] = 24\left[ {3\alpha \beta \gamma - \left( {{\alpha ^3} + {\beta ^3} + {\gamma ^3}} \right)} \right] = 24T$

${\left( {\alpha + \beta + \gamma } \right)^3} = \left( {{\alpha ^3} + {\beta ^3} + {\gamma ^3}} \right) + 3\left[ {\left( {\alpha + \beta + \gamma } \right)\left( {\alpha \beta + \beta \gamma + \alpha \gamma } \right) - \alpha \beta \gamma } \right]$

$ \Rightarrow T = 3\alpha \beta \gamma - \left( {{\alpha ^3} + {\beta ^3} + {\gamma ^3}} \right) = 3\left( {\alpha + \beta + \gamma } \right)\left( {\alpha \beta + \beta \gamma + \alpha \gamma } \right) - {\left( {\alpha + \beta + \gamma } \right)^3}$

$ \Rightarrow T = \left( {\alpha + \beta + \gamma } \right)\left( {3\left( {\alpha \beta + \beta \gamma + \alpha \gamma } \right) - {{\left( {\alpha + \beta + \gamma } \right)}^2}} \right)$

$ \Rightarrow T = \left( {\alpha + \beta + \gamma } \right)\left( {3\left( {\alpha \beta + \beta \gamma + \alpha \gamma } \right) - {{\left( {{\alpha ^2} + {\beta ^2} + {\gamma ^2} + 2\left( {\alpha \beta + \beta \gamma + \alpha \gamma } \right)} \right)}^2}} \right)$

$ \Rightarrow T = \left( {\alpha + \beta + \gamma } \right)\left( {\left( {\alpha \beta + \beta \gamma + \alpha \gamma } \right) - \left( {{\alpha ^2} + {\beta ^2} + {\gamma ^2}} \right)} \right)$

$ \Rightarrow T = - \frac{{\left( {\alpha + \beta + \gamma } \right)}}{2}\left( {2\left( {{\alpha ^2} + {\beta ^2} + {\gamma ^2}} \right) - 2\left( {\alpha \beta + \beta \gamma + \alpha \gamma } \right)} \right)$

$ \Rightarrow T = - \frac{{\left( {\alpha + \beta + \gamma } \right)}}{2}\left( {{{\left( {\alpha - \beta } \right)}^2} + {{\left( {\beta - \gamma } \right)}^2} + {{\left( {\gamma - \alpha } \right)}^2}} \right)$

$ \Rightarrow T = - \left( {a + b + c} \right)\left( {{{\left( {a - c} \right)}^2} + {{\left( {b - a} \right)}^2} + {{\left( {c - b} \right)}^2}} \right)$

$ \Rightarrow T = - 2\left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - \left( {ac + ba + cb} \right)} \right)$

$D \ne 0;a + b + c \ne 0;{a^2} + {b^2} + {c^2} \ne ab + bc + ca$, hence $P \to 3$

Using this how do we find the other options?

Alessio K
  • 10,599
Samar Imam Zaidi
  • 8,800

1 Answers1

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$\bullet$ If $a+b+c=0$ and $ a^2 + b^2 + c^2 \neq ac + ba + cb $ $\implies $ $T=0$ so the equations have infinitely many solutions

$\bullet$ If $a+b+c\neq0$ and $ a^2 + b^2 + c^2 = ac + ba + cb $ $\implies $ $T=0$ and $a=b=c\neq0$ so the equations represent identical planes

$\bullet$ If $a+b+c\neq0$ and $ a^2 + b^2 + c^2 \neq ac + ba + cb $ $\implies $ $T\neq0$ so the equations represent planes meeting only at a single point

$\bullet$ If $a+b+c=0$ and $ a^2 + b^2 + c^2 = ac + ba + cb $ $\implies $ $T=0$ and $a=b=c=0$ so the equations represent the whole of $\mathbb{R^3}$

Alessio K
  • 10,599
  • This is the solution that is there in the book but it is not clear, I only know the proof for unique solution – Samar Imam Zaidi Sep 11 '21 at 09:13
  • Which part is not clear? If $a+b+c=0$ and $a=b=c\neq 0$ (first bullet point), you can substitute into the system of equations and obtain that the equations represent the line $2x=3y=4z$. Then if $T=0$ and $a=b=c=0,$ any $(x,y,z)\in\mathbb{R^3}$ will be a solution to the system. – Alessio K Sep 11 '21 at 09:15
  • i have understood the problem – Samar Imam Zaidi Sep 11 '21 at 09:22