JEE Main 2021 — Three Dimensional Geometry Question with Solution

The given line is $\frac{x+1}{2}=\frac{y-3}{1}=\frac{z+2}{-1}$

$\frac{x+1}{2}=\frac{y-3}{1}=\frac{z+2}{-1}=\lambda ($let$)$

$\Rightarrow \frac{x+1}{2}=\lambda , \frac{y-3}{1}=\lambda , \frac{z+2}{-1}=\lambda$

$\Rightarrow x=2\lambda -1, y=\lambda +3, z=-\lambda -2$

Hence, any point on the line $\frac{x+1}{2}=\frac{y-3}{1}=\frac{z+2}{-1}$ is $\left(2\lambda -1, \lambda +3, -\lambda -2\right).$

Let, $M$ be the foot of perpendicular from the point $P$ to the line $\frac{x+1}{2}=\frac{y-3}{1}=\frac{z+2}{-1},$ hence $M\equiv \left(2\lambda -1, \lambda +3, -\lambda -2\right).$

Now, direction ratios of $\overrightarrow{PM}=(2\lambda -3, \lambda , -\lambda -3)$

Given $\overrightarrow{PM}$ is perpendicular to the line $\frac{x+1}{2}=\frac{y-3}{1}=\frac{z+2}{-1},$ then the sum of the product of their direction ratios is zero.

$2\left(2\lambda -3\right)+1\left(\lambda \right)-1\left(-\lambda -3\right)=0$

$\Rightarrow 4\lambda -6+\lambda +\lambda +3=0\Rightarrow \lambda =\frac{1}{2}$

$\therefore M\equiv \left(0, \frac{7}{2}, -\frac{5}{2}\right)$

The mid-point of the point $P$ and the image of $P$ in the line $\frac{x+1}{2}=\frac{y-3}{1}=\frac{z+2}{-1}$ is $M.$

$\therefore$ Reflection $(-2, 4, -6).$

Thus, the plane passes through the point $(-2, 4, -6).$

Also, the plane contains the line $\frac{x-2}{3}=\frac{1-y}{2}=\frac{z+1}{1}$

$\Rightarrow \frac{x-2}{3}=\frac{y-1}{-2}=\frac{z+1}{1},$ hence it passes through the point $\left(2, 1, -1\right)$ and the normal to the plane is perpendicular to the line.

The equation of a plane passing through the points $\left(x_1, y_1, z_1\right)$ and $\left(x_2, y_2, z_2\right)$ and having normal perpendicular to a line with direction ratios $<a, b, c>$ is $\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ a& b& c\\ x_1-x_2& y_1-y_2& z_1-z_2\end{array}\right|=0$

Thus, the required plane is $\left|\begin{array}{ccc}x-2& y-1& z+1\\ 3& -2& 1\\ 4& -3& 5\end{array}\right|=0$

$\Rightarrow (x-2)(-10+3)-(y-1)(15-4)+(z+1)(-1)=0$

$\Rightarrow -7x+14-11y+11-z-1=0$

$\Rightarrow 7x+11y+z=24$

Given the plane is $\alpha x+\beta y+\gamma z=24$

$\therefore \alpha =7, \beta =11, \gamma =1$

$\Rightarrow \alpha +\beta +\gamma =19.$