Question

The equation of the plane passing through the line $\frac{x-1}{5}=\frac{y+2}{6}=\frac{z-3}{4}$ and the point $(4,3,7)$ is

(a) $4 x+8 y+7 z=41$

(b) $4 x-8 y+7 z=41$

(c) $4 x-8 y-7 z=41$

(d) $4 x-8 y+7 z=39$

Answer 1

Equation of any plane through given line is $a(x-1)+b(y+2)+c(z-3)=0$

Also, passing through $(4,3,7)$

$\therefore  3 a+5 b+4 c=0$

and $5 a+6 b+4 c=0$

$\therefore \frac{a}{-4}=\frac{b}{8}=\frac{c}{-7}$

$\Rightarrow  \frac{a}{4}=\frac{b}{-8}=\frac{c}{7}$

$\therefore$ Required equation is

$4 x-8 y+7 z=41$

So, The correct option of this question will be (B).