Let the line L: x = (1 - y)/(- 2) = (z - 3)/(λ), λ ∈ ℝ meet the...

The given equations are,

$L: \dfrac{x - 0}{1} = \dfrac{y - 1}{2} = \dfrac{z - 3}{\lambda} \text{\text{ and } }$ P: x + 2y + 3z = 4

So, Vector parallel to line$: = \overset{\to}{b}$

Normal vector to plane $P: = \overset{\to}{n}$

Angle between plane & line is $\theta$. Then angle between the line and normal to the plane will be $90 - \theta$.
Then, ⇒ \cos \left(90 - \theta\right) = \dfrac{\left(\overset{^}{i} + 2\overset{^}{j} + \lambda\overset{^}{k}\right) \cdot \left(\overset{^}{i} + 2\overset{^}{j} + 3\overset{^}{k}\right)}{\sqrt{{1}^{2} + {2}^{2} + {\lambda}^{2}} \cdot \sqrt{{1}^{2} + {2}^{2} + {3}^{2}}}
$⇒ \dfrac{3}{\sqrt{14}} = \dfrac{1 + 4 + 3\lambda}{\sqrt{{\lambda}^{2} + 5}\sqrt{14}} ⇒ \lambda = \dfrac{2}{3}$

${L}_{1} = \dfrac{x - 0}{3} = \dfrac{y - 1}{6} = \dfrac{z - 3}{2} = \mu$

\left(x, y, z\right) ≡ \left(3\mu, 6\mu + 1, 2\mu + 3\right)$</p><p>But this point lies on the plane :</p><p>\left(3\mu + 12\mu + 2 + 6\mu + 9\right) = 4$

$⇒ \mu = \dfrac{- 1}{3}$
Hence, $\alpha = 3\mu = - 1, \beta = 6\mu + 1 = - 1, \gamma = 2\mu + 3 = \dfrac{7}{3}$

Now, $\alpha + 2\beta + 6\gamma = 11$

Hence this is the required answer.