Let the image of the point P(1, 6, a) in the line L: x 1 = y-1 2 = z-a+1 b, b > 0, be ( a 3, 0, a+c ). If S(α, β, γ), α > 0, is the point on L such…

Let the foot of the perpendicular from $P(1, 6, a)$P(1,6,a) to the line $L$L be $M$M. Since the image of $P$P is $P'\left(\dfrac{a}{3}, 0, a+c\right)$P′(3a​,0,a+c), $M$M is the midpoint of $P$P and $P'$P′.

$$M = \left(\dfrac{1 + \dfrac{a}{3}}{2}, \dfrac{6 + 0}{2}, \dfrac{a + a + c}{2}\right) = \left(\dfrac{a+3}{6}, 3, a + \dfrac{c}{2}\right)$$M=​21+3a​​,26+0​,2a+a+c​​=(6a+3​,3,a+2c​)

Since $M$M lies on the line $L: \dfrac{x}{1} = \dfrac{y-1}{2} = \dfrac{z-a+1}{b}$L:1x​=2y−1​=bz−a+1​, we substitute its coordinates into the equation of the line:

$$\dfrac{\dfrac{a+3}{6}}{1} = \dfrac{3-1}{2} = \dfrac{a + \dfrac{c}{2} - a + 1}{b}$$16a+3​​=23−1​=ba+2c​−a+1​

$$\dfrac{a+3}{6} = 1 = \dfrac{\dfrac{c}{2} + 1}{b}$$6a+3​=1=b2c​+1​

From $\dfrac{a+3}{6} = 1$6a+3​=1, we get $a = 3$a=3.

From $\dfrac{\dfrac{c}{2} + 1}{b} = 1$b2c​+1​=1, we get $2b = c + 2 \Rightarrow c = 2b - 2$2b=c+2⇒c=2b−2.

The vector $\overrightarrow{PP'}$PP′ is perpendicular to the direction vector of the line $L$L, which is $\vec{v} = \hat{i} + 2\hat{j} + b\hat{k}$v=i^+2j^​+bk^.

$$\overrightarrow{PP'} = \left(\dfrac{a}{3} - 1\right)\hat{i} + (0 - 6)\hat{j} + (a+c - a)\hat{k} = \left(\dfrac{3}{3} - 1\right)\hat{i} - 6\hat{j} + c\hat{k} = -6\hat{j} + c\hat{k}$$PP′=(3a​−1)i^+(0−6)j^​+(a+c−a)k^=(33​−1)i^−6j^​+ck^=−6j^​+ck^

Taking the dot product $\overrightarrow{PP'} \cdot \vec{v} = 0$PP′⋅v=0:

$$0(1) - 6(2) + c(b) = 0 \Rightarrow bc = 12$$0(1)−6(2)+c(b)=0⇒bc=12

Substituting $c = 2b - 2$c=2b−2 into $bc = 12$bc=12:

$$b(2b - 2) = 12 \Rightarrow 2b^2 - 2b - 12 = 0 \Rightarrow b^2 - b - 6 = 0$$b(2b−2)=12⇒2b2−2b−12=0⇒b2−b−6=0

$$(b-3)(b+2) = 0$$(b−3)(b+2)=0

Since $b > 0$b>0, we have $b = 3$b=3. Then $c = 2(3) - 2 = 4$c=2(3)−2=4.

The coordinates of $M$M are $\left(\dfrac{3+3}{6}, 3, 3 + \dfrac{4}{2}\right) = (1, 3, 5)$(63+3​,3,3+24​)=(1,3,5).

The equation of the line $L$L is $\dfrac{x}{1} = \dfrac{y-1}{2} = \dfrac{z-2}{3}$1x​=2y−1​=3z−2​. Any point $S$S on the line can be written as $(t, 2t+1, 3t+2)$(t,2t+1,3t+2).

The distance $SM$SM is given as $2\sqrt{14}$214​.

$$SM^2 = (t-1)^2 + (2t+1-3)^2 + (3t+2-5)^2 = 56$$SM2=(t−1)2+(2t+1−3)2+(3t+2−5)2=56

$$(t-1)^2 + 4(t-1)^2 + 9(t-1)^2 = 56$$(t−1)2+4(t−1)2+9(t−1)2=56

$$14(t-1)^2 = 56 \Rightarrow (t-1)^2 = 4 \Rightarrow t - 1 = \pm 2$$14(t−1)2=56⇒(t−1)2=4⇒t−1=±2

$t = 3$t=3 or $t = -1$t=−1

Since $S(\alpha, \beta, \gamma) = (t, 2t+1, 3t+2)$S(α,β,γ)=(t,2t+1,3t+2) and $\alpha > 0$α>0, we must have $t = 3$t=3.

Thus, $S = (3, 7, 11)$S=(3,7,11).

$$\alpha + \beta + \gamma = 3 + 7 + 11 = 21$$α+β+γ=3+7+11=21