The equation of a plane containing the line $x-2=\frac{y-4}{4}=\frac{z-6}{7}$ and parallel to the line…
The equation of a plane containing the line $x-2=\frac{y-4}{4}=\frac{z-6}{7}$ and parallel to the
line $\bar{r}=(\hat{i}+3 \hat{\jmath}+5 \hat{k})+\lambda(3 \hat{\imath}+5 \hat{\jmath}+7 \hat{k})$ is
$x-2 y+z=10$
$3 x-2 y+z=4$
$x-2 y+z=9$
$x-2 y+z=0$
Solution
d.r. of given liens are $(1,4,7)$ and $(3,5,7)$. Normal to the plane is perpendicular to them.
$\begin{array}{l}
=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 4 & 7 \\
3 & 5 & 7
\end{array}\right| \\
=\hat{i}(-7)-\hat{j}(-14)+\hat{k}(-7) \\
=-7(\hat{i}-2 \hat{j}+\hat{k})
\end{array}$
Hence eq. of plane is $-7(x-2 y+z)=a$
This plane plasses through point $(2,4,6)$
$\therefore \quad a=-7(2-8+6)=0$
$\therefore$ Eq. of plane is $x-2 y+z=0$