If from a point $P (a, b, c)$ perpendiculars $P A, P B$ are drawn to $y z$ and $z x$ planes, then the equation of the plane $O A B$ is

b c x + c a y + a b z = 0

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b c x + c a y - a b z = 0

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b c x - c a y + a b z = 0

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- b c x + c a y + a b z = 0

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Solution

The correct option is B $b c x + c a y - a b z = 0$
$A [0, b, c]$ in $y z$ - plane
$B [a, 0, c]$ in $z x$ - plane.
Plane through $O$ is $p x + q y + r z = 0.$
It passes through $A$ and $B$ .
$∴ 0 p + q b + r c = 0$ and $p a + 0 b + r c = 0$
$∴ \frac{p}{b c} = \frac{q}{c a} = \frac{r}{- a b} .$
Hence the plane is as in ( b ).

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