Question

The line $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ is parallel to the plane.

• A. $2x+3y+4z=29$
• B. $3x+4y-5z=10$
• C. $3x+4y+5z=38$
• D. $x+y+z=0$

Answer 1

The correct option is B $3x+4y-5z=10$

Given line is $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$

D is line are (3,4,5)

Let D is of normal of plane be (a,b,c)

Given that line is parallel to plane

so line is perpendicular to normal of plane

So $3\left(a\right)+4\left(b\right)+5\left(c\right)=0$

so the values of a,b,c satisfying the eq then

is present in $3x+4y-5z=10$

$a=3,b=4,c=-5$