The equation of the plane containing the straight line
Question:

The equation of the plane containing the straight line $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$ and perpendicular to the plane containing the straight lines

$\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$ is:

1. $x+2 y-2 z=0$

2. $x-2 y+z=0$

3. $5 \mathrm{x}+2 \mathrm{y}-4 \mathrm{z}=0$

4. $3 x+2 y-3 z=0$

Correct Option: , 2

Solution:

Vector along the normal to the plane containing the lines

$\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$

is $(8 \hat{\mathrm{i}}-\hat{\mathrm{j}}-10 \hat{\mathrm{k}})$

vector perpendicular to the vectors $2 \hat{i}+3 \hat{j}+4 \hat{k}$ and $8 \hat{\mathrm{i}}-\hat{\mathrm{j}}-10 \hat{\mathrm{k}}$ is $26 \hat{\mathrm{i}}-52 \hat{\mathrm{j}}+26 \hat{\mathrm{k}}$ so, required plane is

$26 x-52 y+26 z=0$

$x-2 y+z=0$