Question

Two intersecting lines lying in plane $P_{1}$ have equations $\frac{x - 1}{1}=\frac{y - 3}{2}=\frac{z - 4}{3}$ and $\frac{x - 1}{2}=\frac{y - 3}{3}=\frac{z - 4}{1}.$ If the equation of plane $P_{2}$ is $7x-5y+z-6=0$ , then the distance between planes $P_{1}$ and $P_{2}$ is

• A. $\frac{11}{5 \sqrt{3}}$
• B. $\frac{2}{\sqrt{3}}$
• C. $\frac{1}{\sqrt{3}}$
• D. $\frac{7}{5 \sqrt{3}}$

Answer 1

Answer: (B)

Equation of plane $P_{1}$ is $\begin{vmatrix} x-1 & y-3 & z-4 \\ 1 & 2 & 3 \\ 2 & 3 & 1 \end{vmatrix}=0$
$\left(x - 1\right)\left(- 7\right)-\left(y - 3\right)\left(- 5\right)+\left(z - 4\right)\left(- 1\right)=0$
$7x-5y+z+4=0$
So, distance between planes $P_{1 }\&P_{2}$ is $=\left|\frac{10}{\sqrt{49 + 25 + 1}}\right|=\frac{10}{5 \sqrt{3}}$
$=\frac{2}{\sqrt{3}}$