The equations of three planes ${{p}_{1}}$, ${{p}_{2}}$ and ${{p}_{3}}$ are $4x+6y-z=3$ \(2x+

Question

The equations of three planes ${{p}_{1}}$, ${{p}_{2}}$ and ${{p}_{3}}$ are $4x+6y-z=3$ $2x+5y-2z=1$ $\alpha x-3y+4z=\beta $ respectively, where $\alpha $ and $\beta $ are constants. Given that $\alpha =5$ and $\beta =9$, find the coordinates of the point that lies on all three planes.[1] Find the values of $\alpha $ and $\beta $ such that the line where ${{p}_{1}}$ and ${{p}_{2}}$ meet lies in ${{p}_{3}}$.[5]

Accepted Answer

(a) $(2,-1,-1)$ (b) $\alpha=2$, $\beta=3$

2023 NJC P1 Q1

2023 NJC P1 Q1

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