If the equation of the plane passing through the line of...

Given,

The equation of the plane passing through the line of intersection of the planes $2x - y + z = 3, 4x - 3y + 5z + 9 = 0$ and parallel to the line $\dfrac{x + 1}{- 2} = \dfrac{y + 3}{4} = \dfrac{z - 2}{5}$ is $ax + by + cz + 6 = 0,$

Now let, ${P}_{1} = 2x - y + z = 3$

And ${P}_{2} = 4x - 3y + 5z + 9 = 0$

Now using family of planes we get,

${P}_{1} + \lambda{P}_{2} = 0$

$⇒ ( 2x - y + z - 3 ) + \lambda ( 4x - 3y + 5z + 9 ) = 0$

$⇒ ( 2 + 4\lambda ) x - ( 1 + 3\lambda ) y + ( 1 + 5\lambda ) z - ( 3 - 9\lambda ) = 0$

Now let,

${P}_{3} = ( 2 + 4\lambda ) x - ( 1 + 3\lambda ) y + ( 1 + 5\lambda ) z - ( 3 - 9\lambda ) = 0$

Now given ${P}_{3}$ is parallel to line,

$\dfrac{x + 1}{- 2} = \dfrac{y + 3}{4} = \dfrac{z - 2}{5}$

So, using perpendicular condition as normal vector of plane will be perpendicular to given line we get,

$- 2 ( 2 + 4\lambda ) - 4 ( 1 + 3\lambda ) + 5 ( 1 + 5\lambda ) = 0$

$⇒ - 3 + 5\lambda = 0$

$⇒ \lambda = \dfrac{3}{5}$

Hence, equation of plane ${P}_{3}$ will be,

${P}_{3}: \dfrac{22x}{5} - \dfrac{14y}{5} + \dfrac{20z}{5} + \dfrac{12}{5} = 0$

$⇒ {P}_{3} = 11x - 7y + 10z + 6 = 0$

Now on comparing with $ax + by + cz + 6 = 0,$ we get,

$a = 11$, $b = - 7$ and $c = 10$

$∴ a + b + c = 11 - 7 + 10 = 14$