Question

The equation of a plane passing through the line of intersection of the planes$x+2y+3z=2,x-y+z=3$ and at a distance $\frac{2}{\sqrt{3}}$ from the point $(3,1,-1)$ is

• A. $5x–11y+z=17$
• B. $\sqrt{2}x+y=3\sqrt{2}–1$
• C. $x+y+z=\sqrt{3}$
• D. $x–\sqrt{2}y=1–\sqrt{2}$

Answer 1

The correct option is A

$5x–11y+z=17$

Explanation for the correct option

Finding the equation of the plane

Equation for the required plane is :

$\begin{array}{rcl}(x+2y+3z-2)+\lambda (x-y+z-3)& =& 0\\ & \Rightarrow & (1+\lambda )x+(2-\lambda )y+(3+\lambda )z-(2+3\lambda )=0\end{array}$

It's distance from $(3,1,-1)$ is $\frac{2}{\sqrt{3}}$.So,

$\begin{array}{lcl}\frac{2}{\sqrt{3}}& =& \frac{\left|3(1+\mathrm{\lambda })+(2-\mathrm{\lambda })-(3+\mathrm{\lambda })-(2+3\mathrm{\lambda })\right|}{\sqrt{{(\mathrm{\lambda }+1)}^2+{(2-\mathrm{\lambda })}^2+{(3+\mathrm{\lambda })}^2}}\\ \Rightarrow \frac{4}{3}& =& \frac{{\left(-2\mathrm{\lambda }\right)}^2}{3{\mathrm{\lambda }}^2+4\mathrm{\lambda }+14}\\ \Rightarrow 3{\mathrm{\lambda }}^2+4\mathrm{\lambda }+14& =& 3{\mathrm{\lambda }}^2\\ \Rightarrow \mathrm{\lambda }& =& -\frac{7}{2}\\ \mathrm{Now},\mathrm{substitue}\mathrm{value}\mathrm{of}\mathrm{\lambda }& & \\ -\frac{5}{2}\mathrm{x}+\frac{11}{2}\mathrm{y}-\frac{\mathrm{z}}{2}+\frac{17}{2}& =& 0\\ \Rightarrow -5\mathrm{x}+11\mathrm{y}-\mathrm{z}+17& =& 0\\ \Rightarrow 5\mathrm{x}-11\mathrm{y}+\mathrm{z}& =& 17\end{array}$

Therefore, the equation of the plane is $5x–11y+z=17$.

Hence, the correct answer is Option (A).