JEE Main 2021 — Three Dimensional Geometry Question with Solution

The given planes are $\overrightarrow{r}·(\hat{i}-\hat{j}+2\hat{k})=2$ and $\overrightarrow{r}·(2\hat{i}+\hat{j}-\hat{k})=2.$

We can convert these planes into cartesian form by taking $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$

Thus, we have $\left(x\hat{i}+y\hat{j}+z\hat{k}\right)·(\hat{i}-\hat{j}+2\hat{k})=2$ and $\left(x\hat{i}+y\hat{j}+z\hat{k}\right)·(2\hat{i}+\hat{j}-\hat{k})=2$

$\Rightarrow x-y+2z=2$ and $2x+y-z=2$

So, let $P_1\equiv x-y+2z=2$ and $P_2\equiv 2x+y-z=2$

Let, the line of Intersection $L$ of planes $P_1$ and $P_2$ cuts $xy$ plane in point $Q$ and it is given that the foot of perpendicular $P$ on $L$ from a point $F$ is $F\left(1, 2, 0\right),$ and this point is also on the $xy$ plane.

From the diagram it is clear that, $PQ$ is a line in the $xy$ plane and hence at every point on it the $z$-coordinate is zero. 

$\Rightarrow z$-coordinate of point $Q,$ is zero and it is on both the planes, 

$\Rightarrow x-y=2 ...\left(i\right)$

And, $2x+y=2 ...\left(ii\right)$

On solving the above two equation, we get $x=\frac{4}{3}, y=\frac{-2}{3}.$

$\Rightarrow Q\left(\frac{4}{3}, -\frac{2}{3}, 0\right)$

A vector parallel to the line of intersection can be obtained by finding a vector which is perpendicular to both the planes.

And, we know that a vector perpendicular to two vectors can be obtained by finding their cross product, i.e. the vector parallel to the line of intersection is

$\overrightarrow{a}=\left|\begin{array}{rrr}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& 2\\ 2& 1& -1\end{array}\right|=-\hat{i}+5\hat{j}+3\hat{k}$

Now, we know that the equation of a line passing through a point $\left(x_1, y_1\right)$ and parallel to a vector $a\hat{i}+b\hat{j}+c\hat{k}$ is $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}.$

Thus, the equation of the line of intersection of the planes is 

$\frac{x-\left({\displaystyle \frac{4}{3}}\right)}{\left(-1\right)}=\frac{y+{\displaystyle \frac{2}{3}}}{5}=\frac{z-0}{3}$

$\frac{x-\left({\displaystyle \frac{4}{3}}\right)}{\left(-1\right)}=\frac{y+{\displaystyle \frac{2}{3}}}{5}=\frac{z}{3}=\lambda$ $($let$)$

$\Rightarrow x=-\lambda +\frac{4}{3}, y=5\lambda -\frac{2}{3} \& z=3\lambda .$

Let, coordinates of foot of perpendicular be

$P\equiv \left(-\lambda +\frac{4}{3}, 5\lambda -\frac{2}{3}, 3\lambda \right)$

Now, $\overrightarrow{PF}=\left(-\lambda +\frac{4}{3}-1\right)\hat{i}+\left(5\lambda -\frac{2}{3}-2\right)\hat{j}+\left(3\lambda -0\right)\hat{k}$

$\Rightarrow \overrightarrow{PF}=\left(-\lambda +\frac{1}{3}\right)\hat{i}+\left(5\lambda -\frac{8}{3}\right)\hat{j}+\left(3\lambda \right)\hat{k}$

Now, since $\overrightarrow{PF}$ is perpendicular to the line, hence the dot product of their direction ratios is zero.

$\Rightarrow \left(-1\right)\left(-\lambda +\frac{1}{3}\right)+5\left(5\lambda -\frac{8}{3}\right)+3\left(3\lambda \right)=0$

$\Rightarrow \lambda -\frac{1}{3}+25\lambda -\frac{40}{3}+9\lambda =0$

$\Rightarrow 35\lambda =\frac{41}{3}$

$\Rightarrow \lambda =\frac{41}{105}.$

Now, $\alpha =-\lambda +\frac{4}{3}, \beta =5\lambda -\frac{2}{3}, \gamma =3\lambda$

$\Rightarrow \alpha +\beta +\gamma =7\lambda +\frac{2}{3}$

$\Rightarrow \alpha +\beta +\gamma =7\left(\frac{41}{105}\right)+\frac{2}{3}$

$\Rightarrow \alpha +\beta +\gamma =\frac{51}{15}$

$\Rightarrow 35(\alpha +\beta +\gamma )=\frac{51}{15}\times 35=119.$