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If from a point $\mathrm{P}(\mathrm{a}, \mathrm{b}, \mathrm{c})$, perpendiculars $\mathrm{PA}$ and $\mathrm{PB}$ are drawn to $\mathrm{YZ}$ and $\mathrm{ZX}$ planes respectively, then the equation of the plane $\mathrm{OAB}$ is
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bcx $+$ cay $-a b z=0$
$\mathrm{A}(\mathrm{o}, \mathrm{b}, \mathrm{c}) \mathrm{B}(\mathrm{a}, \mathrm{o}, \mathrm{c}), \mathrm{O}(\mathrm{o}, \mathrm{o}, \mathrm{o})$ Perpendicular vector is
$\overrightarrow{\mathrm{OA}} \times \overrightarrow{\mathrm{OB}}$
$\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{j} & \hat{k} \\ 0 & b & c \\ a & 0 & c\end{array}\right|=b c \hat{i}+a c \hat{j}-a b \hat{k}$
$\therefore$ Equation of plane is $b c x+a c y-a b z=0$
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