Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

The condition for two lines $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$ to be coplanar is:

$\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|=0$

Substituting the given values $(x_1,y_1,z_1)=(0,2,-3)$ and $(x_2,y_2,z_2)=(2,6,3)$:

$\left|\begin{array}{ccc}2-0& 6-2& 3-(-3)\\ 1& 2& \lambda \\ 2& 3& \lambda \end{array}\right|=0$

$\left|\begin{array}{ccc}2& 4& 6\\ 1& 2& \lambda \\ 2& 3& \lambda \end{array}\right|=0$

Expanding the determinant:

$2(2\lambda -3\lambda )-4(\lambda -2\lambda )+6(3-4)=0$

$-2\lambda +4\lambda -6=0$

$2\lambda =6\Rightarrow \lambda =3$

The direction ratios of the lines are $(1,2,3)$ and $(2,3,3)$. The normal vector to the plane containing these lines is given by the cross product of their direction vectors:

$\vec{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 3\\ 2& 3& 3\end{array}\right|=\hat{i}(6-9)-\hat{j}(3-6)+\hat{k}(3-4)=-3\hat{i}+3\hat{j}-\hat{k}$

The equation of the plane passing through the point $(0,2,-3)$ with normal vector $(-3,3,-1)$ is:

$-3(x-0)+3(y-2)-1(z+3)=0$

$-3x+3y-6-z-3=0$

$3x-3y+z+9=0$

Checking the given options, we substitute the point $(4,10,9)$ into the plane's equation:

$3(4)-3(10)+9+9=12-30+18=0$

Thus, the point $(4,10,9)$ lies on the plane.

Answer: $(4,10,9)$