Let the lines [l_1: (x+5)/(3)=(y+4)/(1)=(z-)/(-2)] and l_2: 3 x+2 y+z-2=0=x-3 y+2 z-13 be …

Question

Let the lines $$l_{1}: \frac{x+5}{3}=\frac{y+4}{1}=\frac{z-\alpha}{-2}$$ and $l_{2}: 3 x+2 y+z-2=0=x-3 y+2 z-13$ be coplanar. If the point $$\mathrm{P}(a, b, c)$$ on $l_{1}$ is nearest to the point $$\mathrm{Q}(-4,-3,2)$$, then $|a|+|b|+|c|$ is equal to

Accepted Answer

$$
\begin{aligned}
& (3 \mathrm{x}+2 \mathrm{y}+\mathrm{z}-2)+\mu(\mathrm{x}-3 \mathrm{y}+2 \mathrm{z}-13)=0 \\
& 3(3+\mu)+1 \cdot(2-3 \mu)-2(1+2 \mu)=0 \\
& 9-4 \mu=0 \\
& \mu=\frac{9}{4} \\
& 4(-15-8+\alpha-2)+9(-5+12+2 \alpha-13)=0 \\
& -100+4 \alpha-54+18 \alpha=0 \\
& \Rightarrow \alpha=7 \\
& \text { Let } \mathrm{P} \equiv(3 \lambda-5, \lambda-4,-2 \lambda+7) \\
& \text { Direction ratio of } \mathrm{PQ}(3 \lambda-1, \lambda-1,-2 \lambda+5) \\
& \text { But PQ } \perp \ell_1 \\
& \Rightarrow 3(3 \lambda-1)+1 \cdot(\lambda-1)-2(-2 \lambda+5)=0 \\
& \Rightarrow \lambda=1 \\
& \mathrm{P}(-2,-3,5) \Rightarrow|\mathrm{a}|+|\mathrm{b}|+|\mathrm{c}|=10
\end{aligned}
$$

Solutions · 1