Question

Consider the plane $\pi :x+y=z$, point A(1, 2, -3) and a line $L:\frac{x-1}{3}=\frac{y-2}{-1}=\frac{z-3}{4}$

The equation of the plane containing the line L and the point A is

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

Answer 1

The correct option is D

x + 3y = 7

$B=(3\lambda +1,-\lambda +2,4\lambda +3)1\left(3\lambda \right)+1(-\lambda )-1(4\lambda +6)=0\lambda =-3B=(-8,5,-9)$
Equation is $a(x-1)+b(y-2)+c(z-3)=0$
It passes through $(1,2,-3)\Rightarrow c=0$
and $3a-b=0\Rightarrow a=1andb=3$
$(x-1)+3(y-2)=0$