Let ( α, β, γ ) be the foot of perpendicular from the point ( 1, 2,...

Given: $P\left(\alpha, \beta, \gamma\right)$ be the foot of the perpendicular from the point $Q\left(1, 2, 3\right)$ on the line $\dfrac{x + 3}{5} = \dfrac{y - 1}{2} = \dfrac{z + 4}{3}$ 

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

So, any point on the line is given by \left(5\lambda - 3, 2\lambda + 1, 3\lambda - 4\right)$, so let the point be$P\left(5\lambda - 3, 2\lambda + 1, 3\lambda - 4\right)$</p><p>Now, direction ratio of$PQ = \left(5\lambda - 4, 2\lambda - 1, 3\lambda - 7\right)$</p><p>Now, the line and$PQ$will be perpendicular,</p><p>So,$5\left(5\lambda - 4\right) + 2\left(2\lambda - 1\right) + 3\left(3\lambda - 7\right) = 0$</p><p>$⇒ 38\lambda - 43 = 0$</p><p>$⇒ \lambda = \dfrac{43}{38}$</p><p>So, the point$P\left(\alpha, \beta, \gamma\right) = P\left(5 \times \dfrac{43}{38} - 3, 2 \times \dfrac{43}{38} + 1, 3 \times \dfrac{43}{38} - 4\right)$</p><p>Then$19\left(\alpha + \beta + \gamma\right) = 19\left[5 \times \dfrac{43}{38} - 3 + 2 \times \dfrac{43}{38} + 1 + 3 \times \dfrac{43}{38} - 4\right]$</p><p>$⇒ 19\left(\alpha + \beta + \gamma\right) = 19\left[\dfrac{101 + 124 - 23}{38}\right]$</p><p>$⇒ 19\left(\alpha + \beta + \gamma\right) = 101$