<s>[INST] PLANE TYPE COLUMN BASED QUESTION 
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?
 [/INST] $\bullet$ 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 $\mathbb{R^3}$
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