Question

Consider a plane $x+y-z=1$ and point $A(1,2,-3).$ A line $L$ has the equation $x=1+3r,y=2-r$ and $z=3+4r.$

The equation of the plane containing line $L$ and point $A$ has the equation

• A. $x-3y+5=0$
• B. $x+3y-7=0$
• C. $3x-y-1=0$
• D. $3x+y-5=0$

Answer 1

The correct option is B $x+3y-7=0$

The equation of plane containing the line $L$ is,
$A(x-1)+B(y-2)+C(z-3)=0,$
where $3A-B+4C=0...\left(1\right)$

Eqn $\left(1\right)$ also contains point $A(1,2,-3).$
Hence, $C=0$ and $3A=B$

The equation of plane is,
$x-1+3(y-2)=0$
$\Rightarrow x+3y-7=0$