JEE Main 2017 — Three Dimensional Geometry Question with Solution

Direction ratio's of the plane perpendicular to two given lines is, $\overrightarrow{n}= {\overrightarrow{n}}_1\times {\overrightarrow{n}}_2$

$=\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 6& 7& 8\\ 3& 5& 7\end{array}\right|$

$=\hat{\mathrm{i}}\left(49-40\right)-\hat{\mathrm{j}}\left(42-24\right)+\hat{\mathrm{k}}\left(30-21\right)$

$=9\hat{\mathrm{i}}-18\hat{\mathrm{j}}+9\hat{\mathrm{k}}$

Hence, $dr\text{'}s=\left(9,-18,9\right)$

Now, plane passes through $\left(-1,1, 3\right)$ (A point on the first line),

Hence, the equation of plane containing lines is

$1\left(x+1\right)-2\left(y-1\right)+\left(\mathrm{z}-3\right)=0$

$\Rightarrow x-2y+z=0$

$\therefore$ Foot of perpendicular of point $\left(1,-2,1\right)$ is

$\frac{x-1}{1}=\frac{y+2}{-2}=\frac{z-1}{1}= -\frac{\left[1+4+1\right]}{6}=-1$

$x-1=-1 , y+2=2, z-1=-1$

$x=0, y=0, z=0$