In R ^ 3 , consider the planes P _ 1 : y = 0 and P _ 2 : x + z = 1...

{R}^{3}

, consider the planes

{P}_{1}: y = 0 and {P}_{2}: x + z = 1 .

Let

{P}_{3}

, be a plane, different from

{P}_{1} and {P}_{2}

, which passes through the intersection of

{P}_{1} and {P}_{2}

. If the distance of the point \left(0, 1, 0\right)

, from

{P}_{3}

, is

, and the distance of a point

( \alpha, \beta, \gamma )

, from

{P}_{3}

, is

2$, then which of the following relations is (are) true ? In R ^ 3 , consider the planes P _ 1 : y = 0 and P _ 2 : x + z = 1 . Let P _ 3 , be a plane, different from P _ 1 and P

Options

A $2\alpha + \beta + 2\gamma + 2 = 0$
B $2\alpha - \beta + 2\gamma + 4 = 0$
C $2\alpha + \beta - 2\gamma - 10 = 0$
D $2\alpha - \beta + 2\gamma - 8 = 0$

Correct answer

B. $2\alpha - \beta + 2\gamma + 4 = 0$

Step-by-step solution

${P}_{1}: y = 0$

${P}_{2}: x + z = 1$

${P}_{3}: y + \lambda\left(x + z - 1\right) = 0$

In R ^ 3 , consider the planes P _ 1 : y = 0 and P _ 2 : x + z = 1 . Let P _ 3 , be a plane, different from P _ 1 and P
$∴ \left|\dfrac{\beta + \lambda\left(\alpha + \gamma - 1\right)}{\sqrt{1 + 2{\lambda}^{2}}}\right| = \pm 2$

$∴ \beta + \lambda\left(\alpha + \gamma - 1\right) = \pm 2\sqrt{1 + 2{\lambda}^{2}}$

In R ^ 3 , consider the planes P _ 1 : y = 0 and P _ 2 : x + z = 1 . Let P _ 3 , be a plane, different from P _ 1 and P
$\dfrac{1 - \lambda}{\sqrt{1 + 2{\lambda}^{2}}} = \pm 1$
$\lambda = - 2\left(\lambda \ne 0\right)$

$∴ \beta - 2\left(\alpha + \gamma - 1\right) = \pm 2\left(3\right)$

$2\alpha - \beta + 2\gamma + 4 = 0$ or $2\alpha - \beta + 2\gamma - 8 = 0$ .

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