The point of intersection C of the plane 8x + y + 2z = 0 and the...

Given,

Plane: $8x + y + 2z = 0$

Now equation of line $AB: \dfrac{x - 2}{5} = \dfrac{y - 4}{10} = \dfrac{z + 3}{- 4} = \lambda$

Any point on line $AB$ is given by,

\left(5\lambda + 2, 10\lambda + 4, - 4\lambda - 3\right)$</p><p>Now using perpendicular condition to find the point of intersection of line and plane,$8\left(5\lambda + 2\right) + 10\lambda + 4 - 8\lambda - 6 = 0$</p><p>$⇒ \lambda = - \dfrac{1}{3}$</p><p>So, point of intersection will be$C\left(\dfrac{1}{3}, \dfrac{2}{3}, - \dfrac{5}{3}\right)$</p><p>Now let$L: \dfrac{x - 1}{- 1} = \dfrac{y + 4}{2} = \dfrac{z + 2}{3} = \mu$</p><p>Now any point on the line$L$is given by,</p><p>\left(- \mu + 1, 2\mu - 4, 3\mu - 2\right)$

Now direction ratio of \overset{\to}{CD} = \left(- \mu + \dfrac{2}{3}\right)\overset{^}{i} + \left(2\mu - \dfrac{14}{3}\right)\overset{^}{j} + \left(3\mu - \dfrac{1}{3}\right)\overset{^}{k}

Now again using the perpendicular condition we get,

\left(- \mu + \dfrac{2}{3}\right) ( - 1 ) + \left(2\mu - \dfrac{14}{3}\right)2 + \left(3\mu - \dfrac{1}{3}\right)3 = 0$</p><p>$⇒ \mu = \dfrac{11}{14}$</p><p>Hence,$\overset{\to}{CD} = \dfrac{- 5}{42}, \dfrac{- 130}{42}, \dfrac{85}{42}$</p><p>Now on simplifying the Direction ratios we get, \left(\dfrac{- 5}{42}, \dfrac{- 130}{42}, \dfrac{85}{42}\right) ≡ \left(- 1, - 26, 17\right)$

Now on comparing with \left(a, b, c\right)$we get, \left|a + b + c\right| = 10$.