Question

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

• A. $3x+4y+5z=7$
• B. $2x+y-2z=0$
• C. $x+y-z=2$
• D. $2x+3y+4z=0$

Answer 1

Answer: (B)

$2x+y-2z=0$

Given,

Line $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$

Since,

the coefficients are $(3,4,5)$

the equation of plane should satisfy the points,

here, we use trial and error method, let's check for $2x+y-2z=0$

$=2\left(3\right)+4\left(4\right)-2\left(5\right)$

$=6+3-10$

$=0$

Hence the equation of plane parallel to $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ is $2x+y-2z=0$