If the lines $3 x+y-2=0, p x+2 y-3=0$ and $2 x-y-3=0$ are concurrent, then $p=$

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Aicharya · Accepted Answer

It is given that lines $3 x+y-2=0, p x+2 y-3=0$ and $2 x-y-3=0$ are concurrent, so $\begin{array}{lll} \left|\begin{array}{rrr} 3 & 1 & -2 \ p & 2 & -3 \ 2 & -1 & -3 \end{array}\right|=0 \ \Rightarrow 3(-6-3)-1(-3 p+6)-2(-p-4)=0 \ \Rightarrow -27+3 p-6+2 p+8=0 \ \Rightarrow 5 p-25=0 \Rightarrow p=5 \end{array}$ Hence, option (b) is correct.

If the lines \(3 x+y-2=0, p x+2 y-3=0\) and \(2 x-y-3=0\) are concurrent, then \(p=\)

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