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:
Correct Option: , 2
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
$26 x-52 y+26 z=0$
$x-2 y+z=0$
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