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__Hint:__The position of the point $P({{x}_{1}},{{y}_{1}},{{z}_{1}})$ with respect to given plane $ax+by+cz=p$ is determined by the sign of expression $a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}-p$. If two points $P, Q$ lie on the opposite side of the plane then the product of their respective expressions should be less than $0$ because when the point lies on the given plane it satisfy equation of the plane and if it doesn’t lie on the given plane, the point’s intercept $(x,y,z)$ when substituted in the equation of the plane, gives us a number either positive or negative. Now if we have two points that lie on the same side of the plane, they will give us the same sign number and their respective product will be positive and if they lie on the opposite side the product will be negative.

__Complete step by step answer:__

We have following points and the given plane:

$P(1,2,3)$

$Q(2,-1,0)$

$2x+3y-2z=k..................(i)$

We also know that the points lie on opposite sides of the given plane.

Now, converting the equation (i) to $a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}-p$ form.

Which gives us, $2x+3y-2z-k...............(ii)$

Putting the values of $x,y,z$ intercepts of points $P$ in equation (ii), we get:

$A=2(1)+3(2)-2(3)-k$

$A=2+6-6-k$

$A=6-k...................(iii)$

Now, putting the values of $x,y,z$ intercepts of points $Q$ in equation (ii), we get:

$B=2(2)+3(-1)-2(0)-k$

$B=4-3-0-k$

$B=1-k...................(iv)$

Multiplying $A$ and $B$ and equating the product to less than $0$ because the product of both the expressions must be less than zero, we get:

$A\times B<0$

$(2-k)\times (1-k)<0$

Rearranging the expression, we get:

$[-(k-2)]\times [-(k-1)]<0$

$(k-2)\times (k-1)<0................(v)$

The equation (v) is of inequality form and we need to do trails one by one from the given options.

Now, looking to the equation (v) we can see, if $k<1$ then the multiplication will generate a positive value and the equation (v) becomes invalid.

Again, if $k>2$ then the multiplication will again generate a positive value and the equation (v) again becomes invalid.

Now, if $k$ is of the range 1$<$k$<$2 the equation (v) becomes valid as the multiplication results in negative value.

Hence, the final answer is option (d).

__Note:__

The chances of making mistakes in interpreting results while looking to $(k-2)\times (k-1)<0$ are there. While interpreting the final answer the shortcut key is to use the options and try to interpret the results when there is a need for hit and trial methods.

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