If the straight lines $\dfrac{x-1}{1}=\dfrac{y-2}{2}=\dfrac{z-3}{m^2}$ and $\dfrac{x-3}{1}=\dfrac{y-2}{m^2}=\dfrac{z-1}{2}$ are coplanar, find the distinct real values of $m$.

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Accepted Answer

> [!TLDR]
>
> Set up the coplanarity determinant with $m^2$ as an unknown entry, expand it into a quartic that's really quadratic in $m^2$, and discard the negative root of $m^2$.

**Step 1. Data.** $\vec a=(1,2,3),\ \vec b=(1,2,m^2);\quad \vec c=(3,2,1),\ \vec d=(1,m^2,2)$.

**Step 2. Coplanarity determinant.** $\vec c-\vec a=(2,0,-2)$.
$$\begin{vmatrix}2&0&-2\1&2&m^2\1&m^2&2\end{vmatrix}=0.$$

**Step 3. Expand along row 1.**
$$2\begin{vmatrix}2&m^2\m^2&2\end{vmatrix}-0+(-2)\begin{vmatrix}1&2\1&m^2\end{vmatrix}=2(4-m^4)-2(m^2-2)=8-2m^4-2m^2+4=12-2m^4-2m^2.$$ …

Q.If the straight lines $\dfrac{x-1}{1}=\dfrac{y-2}{2}=\dfrac{z-3}{m^2}$ and $\dfrac{x-3}{1}=\dfrac{y-2}{m^2}=\dfrac{z-1}{2}$ are coplanar, find the distinct real values of $m$ .

Concept understanding — Equation of a Plane

Equation of a Plane

Point + normal form

Normal (perpendicular) form

Intercept form …

Q.If the straight lines x−11=y−22=z−3m2\dfrac{x-1}{1}=\dfrac{y-2}{2}=\dfrac{z-3}{m^2}1x−1​=2y−2​=m2z−3​ and x−31=y−2m2=z−12\dfrac{x-3}{1}=\dfrac{y-2}{m^2}=\dfrac{z-1}{2}1x−3​=m2y−2​=2z−1​ are coplanar, find the distinct real values of mmm.

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Q.If the straight lines $\dfrac{x-1}{1}=\dfrac{y-2}{2}=\dfrac{z-3}{m^2}$ and $\dfrac{x-3}{1}=\dfrac{y-2}{m^2}=\dfrac{z-1}{2}$ are coplanar, find the distinct real values of $m$ .

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