Question

Find the equation of the plane passing through the straight line $\frac{x-1}{2}=\frac{y+2}{-3}=\frac{z}{5}$ and perpendicular to the plane $x-y+z+2=0$

• A. $2x+3y+z+4=0$
• B. $3x+2y+z+6=0$
• C. $2x+3y+4z-2=0$
• D. $3x+4y+2z+7=0$

Answer 1

The correct option is A $2x+3y+z+4=0$

$L:\frac{x-1}{2}=\frac{y+2}{-3}=\frac{z-0}{5}$

$\therefore$ The plane passes through $(1,-2,0)$

Let the equation of the plane is $a(x-1)+b(y-2)+c(z-0)=0$

Since the line L is on the plane so $2a-3b+5c=0$ . . . eq.1

Again the plane is perpendicular to the plane $x-y+z+2=0$ So

$a-b+c=0$ . . . eq.2

Solving eq.1 & eq.2 using cross-multiplication method,

$\frac{a}{(-3)-(-5)}=\frac{b}{5-2}=\frac{c}{(-2)-(-3)}\Rightarrow \frac{a}{2}=\frac{b}{3}=\frac{c}{1}=k\left(say\right),(\ne =0)a=2k,b=3k,c=k.$

Putting these on the equation of the required plane,

$2k(x-1)+3k(y+2)+k(z-0)=0\Rightarrow 2x-2+3y+6+z=0\Rightarrow 2x+3y+z+4=0$