Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The given lines are:
$\bar{r}=(4\hat{j}-\hat{k})+\lambda (\hat{i}+2\hat{j}-2\hat{k})$
$\bar{r}=(2\hat{i}+\hat{j})+\mu (\hat{i}+2\hat{j}-2\hat{k})$

The lines are parallel with direction vector $\vec{b}=\hat{i}+2\hat{j}-2\hat{k}$.
Let $A(0,4,-1)$ be a point on the first line and $B(2,1,0)$ be a point on the second line.
The vector joining $A$ and $B$ is $\overrightarrow{AB}=2\hat{i}-3\hat{j}+\hat{k}$.

The normal vector $\vec{n}$ to the plane containing these two lines is perpendicular to both $\vec{b}$ and $\overrightarrow{AB}$:
$\vec{n}=\vec{b}\times \overrightarrow{AB}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& -2\\ 2& -3& 1\end{array}\right|$
$\vec{n}=\hat{i}(2-6)-\hat{j}(1-(-4))+\hat{k}(-3-4)=-4\hat{i}-5\hat{j}-7\hat{k}$

Taking the normal vector as $4\hat{i}+5\hat{j}+7\hat{k}$, the equation of the plane passing through $A(0,4,-1)$ is:
$4(x-0)+5(y-4)+7(z+1)=0$
$4x+5y+7z-13=0$

The perpendicular distance $d$ of the point $P(2,5,10)$ from this plane is:
$d=\frac{|4(2)+5(5)+7(10)-13|}{\sqrt{4^2+5^2+7^2}}$
$d=\frac{|8+25+70-13|}{\sqrt{16+25+49}}$
$d=\frac{90}{\sqrt{90}}=\sqrt{90}$

Therefore, $d^2=90$.

Answer: $90$