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

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(B) The coordinates of point $P$ are $(a, b, c)$.Since $PA$ is perpendicular to the $yz$-plane,the coordinates of $A$ are $(0, b, c)$.Since $PB$ is pe

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$bcx + cay - abz = 0$

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(B) The coordinates of point $P$ are $(a, b, c)$.Since $PA$ is perpendicular to the $yz$-plane,the coordinates of $A$ are $(0, b, c)$.Since $PB$ is perpendicular to the $zx$-plane,the coordinates of $B$ are $(a, 0, c)$.The origin $O$ is $(0, 0, 0)$.The equation of a plane passing through the origin is $px + qy + rz = 0$.Since the plane passes through $A(0, b, c)$,we have $qb + rc = 0 \Rightarrow qb = -rc$.Since the plane passes through $B(a, 0, c)$,we have $pa + rc = 0 \Rightarrow pa = -rc$.From these,$pa = qb = -rc = k$ (let).Then $p = k/a$,$q = k/b$,and $r = -k/c$.Substituting these into the plane equation: $(k/a)x + (k/b)y - (k/c)z = 0$.Multiplying by $abc/k$,we get $bcx + acy - abz = 0$ is incorrect based on the derivation; let's re-evaluate.Using the ratio method: $p/bc = q/ca = r/(-ab) = k$.Then $p = bck, q = cak, r = -abk$.The equation is $bcx + cay - abz = 0$.

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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