If the distances of the point (1,2,a) from the line L: x-11=y2=z-11, measured along the li…

Question

If the distances of the point $(1,2,a)$ from the line $L:\ \dfrac{x-1}{1}=\dfrac{y}{2}=\dfrac{z-1}{1}$, measured along the lines $L_1:\ \dfrac{x-1}{3}=\dfrac{y-2}{4}=\dfrac{z-a}{b}$ and $L_2:\ \dfrac{x-1}{1}=\dfrac{y-2}{4}=\dfrac{z-a}{c}$ (both through $(1,2,a)$, meeting $L$), are equal, then $a+b+c$ is equal to:

Accepted Answer

Let $P=(1,2,a)$. A general point on $L$ is $Q(t+1,\ 2t,\ t+1)$.

Along $L_1$: $\vec{PQ_1}=(t_1,\ 2t_1-2,\ t_1+1-a)$ must be parallel to $(3,4,b)$:
$$\frac{t_1}{3}=\frac{2t_1-2}{4}\implies 4t_1=6t_1-6\implies t_1=3.$$
Then $\dfrac{3}{3}=\dfrac{3-a+1}{b}\implies a+b=4.\qquad(1)$
$$d_1^2=t_1^2+(2t_1-2)^2+(t_1+1-a)^2=9+16+(4-a)^2=a^2-8a+41.$$

Along $L_2$: $\vec{PQ_2}=(t_2,\ 2t_2-2,\ t_2+1-a)$ parallel to $(1,4,c)$:
$$\frac{t_2}{1}=\frac{2t_2-2}{4}\implies 4t_2=2t_2-2\implies t_2=-1.$$
Then $\dfrac{-1}{1}=\dfrac{-1+1-a}{c}\implies a=c.\qquad(2)$
$d_2^2=1+16+a^2=a^2+17.$

Setting $d_1^2=d_2^2$: $a^2-8a+41=a^2+17\implies 8a=24\implies a=3.$
Then $b=4-a=1$, $c=a=3$.
$a+b+c=3+1+3=7.$

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