# Value of x for which $(x,x+1,x+2)$ and $(1-x,x+2,x)$ lies on opposite side of plane  $x+y+z-2=0$ is Option 1) $x\in (\frac{1}{3},1)$ Option 2) $x\in (\frac{-1}{3},1)$ Option 3) $x\in (-1,\frac{-1}{3})$ Option 4) $x\in (-1,\frac{1}{3})$

H Himanshu

As we have learned

Two sides of plane -

Let the plane be   $ax+by+cz+d= 0$

and two points be $A(x_{1},y_{1},z_{1}),B(x_{2},y_{2},z_{2})$ then

If they are on same side:

$\left ( ax_{1}+by_{1}+cz_{1}+d \right )\left ( ax_{2}+by_{2}+cz_{2}+d \right )> 0$

If they are on opposite side:

$\left ( ax_{1}+by_{1}+cz_{1}+d \right )\left ( ax_{2}+by_{2}+cz_{2}+d \right )< 0$

-

On putting points in L.H.S of equation of plane one by one, we get expression

(x)+(x+1)+(x+2)-2=3x+1

and (1-x)+(x+2)+(x-2)=x+1

For points in opposite side of plane,

$(3x+1)(x+1)< 0$

$\Rightarrow x\in (-1,\frac{-1}{3})$

Option 1)

$x\in (\frac{1}{3},1)$

Option 2)

$x\in (\frac{-1}{3},1)$

Option 3)

$x\in (-1,\frac{-1}{3})$

Option 4)

$x\in (-1,\frac{1}{3})$

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