Let a, b, c are the direction ratios of the normal and (x, y, z) be the co-ordinate of the point through which the plane passes. Then the equation of plane is:

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Solution

The correct option is B

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= ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{1}{(a + x)} & \frac{1}{(b + x)} & \frac{1}{(c + x)} \frac{1}{(a + y)} & \frac{1}{(b + y)} & \frac{1}{(c + y)} \frac{1}{(a + z)} & \frac{1}{(b + z)} & \frac{1}{(c + z)} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ = \frac{P}{Q}

(x_{1}, y_{1}, z_{1})

a, b, c

The line $\frac{x - x_{1}}{0} = \frac{y - y_{1}}{1} = \frac{z - z_{1}}{2}$ is:

x = \frac{a - b}{a + b}, y = \frac{b - c}{b + c}, z = \frac{c - a}{c + a}

\frac{(1 + x) (1 + y) (1 + z)}{(1 - x) (1 - y) (1 - z)}

x = \frac{a - b}{a + b}, y = \frac{b - c}{b + c}

z = \frac{c - a}{c + a}

\frac{(1 + x) (1 + y) (1 + z)}{(1 - x) (1 - y) (1 - z)}

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