If from a point P(a, b, c), perpendiculars PA | vedclass

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(B) Given point $P(a, b, c)$.Perpendicular $PA$ is drawn to the $YZ$-plane. The coordinates of $A$ are $(0, b, c)$.Perpendicular $PB$ is drawn to the

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$bcx + acy - abz = 0$

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(B) Given point $P(a, b, c)$.Perpendicular $PA$ is drawn to the $YZ$-plane. The coordinates of $A$ are $(0, b, c)$.Perpendicular $PB$ is drawn to the $ZX$-plane. The coordinates of $B$ are $(a, 0, c)$.The origin $O$ is $(0, 0, 0)$.The plane passes through $O(0, 0, 0)$, $A(0, b, c)$, and $B(a, 0, c)$.The normal vector $\vec{n}$ to the plane is given by $\vec{OA} 	imes \vec{OB}$.$\vec{OA} = 0\hat{i} + b\hat{j} + c\hat{k}$$\vec{OB} = a\hat{i} + 0\hat{j} + c\hat{k}$$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & b & c \ a & 0 & c \end{vmatrix} = \hat{i}(bc - 0) - \hat{j}(0 - ac) + \hat{k}(0 - ab) = bc\hat{i} + ac\hat{j} - ab\hat{k}$.The equation of the plane passing through the origin is $bcx + acy - abz = 0$.

If from a point $P(a, b, c)$, perpendiculars $PA$ and $PB$ are drawn to $YZ$ and $ZX$ planes respectively, then the equation of the plane $OAB$ is

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