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# The number of distinct real values of $\lambda$ for which the lines $\dfrac{{x - 1}}{1} = \dfrac{{y - 2}}{2} = \dfrac{{z + 3}}{{{\lambda ^2}}}$and$\dfrac{{x - 3}}{1} = \dfrac{{y - 2}}{{{\lambda ^2}}} = \dfrac{{z - 1}}{2}$ are coplanar is:A.$2$B.$4$C.$3$D.$1$

Last updated date: 03rd Mar 2024
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Hint: In this problem, we need to find the coplanar from these two lines of numbers of distinct real values. Lines in the same plane are coplanar lines. Skew lines are lines that do not intersect, and there is no plane that contains them. Intersecting lines are two coplanar lines with exactly one point in common. Concurrent lines are lines that contain the same point.
$\left| {\begin{array}{*{20}{c}} {{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}} \\ {{a_1}}&{{b_1}}&{{c_1}} \\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right|$
Since, the coplanar lines are ${L_1} = \dfrac{{x - {x_1}}}{{{a_1}}} = \dfrac{{y - {y_1}}}{{{b_1}}} = \dfrac{{z - {z_1}}}{{{c_1}}}$ and ${L_2} = \dfrac{{x - {x_2}}}{{{a_2}}} = \dfrac{{y - {y_2}}}{{{b_2}}} = \dfrac{{z - {z_2}}}{{{c_2}}}$

We are given the equation of lines are $\dfrac{{x - 1}}{1} = \dfrac{{y - 2}}{2} = \dfrac{{z + 3}}{{{\lambda ^2}}}$ and $\dfrac{{x - 3}}{1} = \dfrac{{y - 2}}{{{\lambda ^2}}} = \dfrac{{z - 1}}{2}$
The number of distinct has the real values of $\lambda$ for the two lines, we get
Let us consider the two coplanar lines as ${L_1}$and ${L_2}$.
${L_1} = \dfrac{{x - 1}}{1} = \dfrac{{y - 2}}{2} = \dfrac{{z + 3}}{{{\lambda ^2}}}$----------(1)
${L_2} = \dfrac{{x - 3}}{1} = \dfrac{{y - 2}}{{{\lambda ^2}}} = \dfrac{{z - 1}}{2}$----------(2)
Comparing the coplanar equation (1) and (2) with the following line formula:
${L_1} = \dfrac{{x - {x_1}}}{{{a_1}}} = \dfrac{{y - {y_1}}}{{{b_1}}} = \dfrac{{z - {z_1}}}{{{c_1}}}$
${L_2} = \dfrac{{x - {x_2}}}{{{a_2}}} = \dfrac{{y - {y_2}}}{{{b_2}}} = \dfrac{{z - {z_2}}}{{{c_2}}}$
We have the formula for finding the coplanar:
$\left| {\begin{array}{*{20}{c}} {{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}} \\ {{a_1}}&{{b_1}}&{{c_1}} \\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right| = 0$
Since ${x_1} = 1,{y_1} = 2,{z_1} = - 3$,${x_2} = 3,{y_1} = 2,{z_1} = 1$ and$({a_1},{b_1},{c_1}) = (1,2,{\lambda ^2})$, $({a_2},{b_2},{c_2}) = (1,{\lambda ^2},2)$
Here, we have to substitute all the values in coplanar formula, then
$\left| {\begin{array}{*{20}{c}} {3 - 1}&{2 - 2}&{1 - ( - 3)} \\ 1&2&{{\lambda ^2}} \\ 1&{{\lambda ^2}}&2 \end{array}} \right| = 0$
Expanding the last element of first row in further simplification, we can get
$\left| {\begin{array}{*{20}{c}} 2&0&4 \\ 1&2&{{\lambda ^2}} \\ 1&{{\lambda ^2}}&2 \end{array}} \right| = 0$
We do perform the determinant operation simplified as follows, we get
$2(4 - {\lambda ^4}) - 0(2 - {\lambda ^2}) + 4({\lambda ^2} - 2) = 0 \\ 2(4 - {\lambda ^4}) + 4({\lambda ^2} - 2) = 0 \\$
Now, we have to simplify it, dividing the equation by $2$, we get
$\Rightarrow (4 - {\lambda ^4}) + 2({\lambda ^2} - 2) = 0$
Expanding the brackets to simplify in further:
$\Rightarrow 4 - {\lambda ^4} + 2{\lambda ^2} - 4 = 0 \\ \Rightarrow - {\lambda ^4} + 2{\lambda ^2} = 0 \\$
Here, we take common factors out, we can get
$- {\lambda ^2}({\lambda ^2} - 2) = 0$
$\lambda = 0, \pm \sqrt 2$
Therefore, $\lambda = 0,\sqrt 2 , - \sqrt 2$. So, the coplanar is $3$.
Finally, the correct answer is Option (C) $3$
As a result, The number of distinct real values of $\lambda$ for which the lines $\dfrac{{x - 1}}{1} = \dfrac{{y - 2}}{2} = \dfrac{{z + 3}}{{{\lambda ^2}}}$ and $\dfrac{{x - 3}}{1} = \dfrac{{y - 2}}{{{\lambda ^2}}} = \dfrac{{z - 1}}{2}$ are coplanar is $3$.
So, the correct answer is “Option C”.

Note: We note that the two lines are coplanar lies on the plane. When two lines lie on the same plane in three dimensions, they are assumed to be coplanar. We've learned how to use vector notations to describe a line's equation in three dimensions. If any three points determine a plane then additional points can be checked for coplanar by measuring the distance of the points from the plane, if the distance is zero then the point is coplanar.