# If the points (1, 1, ) and (-3, 0, 1) are equidistant from the plane, 3x+4y-12z+13=0, then satisfies the equation : Option 1) Option 2) Option 3) Option 4)

V Vakul

As we learnt in

Distance of a point from plane (Cartesian form) -

The length of perpendicular from $P(x_{1},y_{1},z_{1})$ to the plane

$ax+by+cz+d= 0$ is given by  $\frac{\left [ ax_{1}+by_{1} +cz_{1}+d\right ]}{\left | \sqrt{a^{2}+b^{2}+c^{2}} \right |}$

-

Distance from plane are

$\left | \frac{3+4-12\lambda+13}{13} \right |$$=\left | \frac{-9-12+13}{13} \right |$

$\left | 20-12\lambda \right |=8$

$\left | 3\lambda-5 \right |=2$

$3\lambda-5=2$         or         $3\lambda-5=-2$

$\lambda=\frac{7}{3}$                    or            $\lambda=1$

These are roots of equation

$3x^{2}-10x+7=0$

Option 1)

This option is correct

Option 2)

This option is incorrect

Option 3)

This option is incorrect

Option 4)

This option is incorrect

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