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

Question

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

Accepted Answer

Family for $\ell_2$'s planes: $(3x+2y+z-2)+\mu(x-3y+2z-13)=0$. Coplanarity with $\ell_1$'s direction $(3,1,-2)$ requires this plane's normal $\cdot$ direction $=0$:
$$3(3+\mu)+(2-3\mu)-2(1+2\mu)=0 \implies 9-4\mu=0 \implies \mu=\frac94.$$
Using the point condition on $\ell_1$ and the combined plane to solve for $\alpha$:
$$4(-15-8+\alpha-2)+9(-5+12+2\alpha-13)=0 \implies -100+4\alpha-54+18\alpha=0 \implies \alpha=7.$$

$P=(3\lambda-5,\ \lambda-4,\ -2\lambda+7)$ on $\ell_1$; $\vec{PQ}=(3\lambda-1,\ \lambda-1,\ -2\lambda+5)$, perpendicular to $\ell_1$'s direction:
$$3(3\lambda-1)+(\lambda-1)-2(-2\lambda+5)=0 \implies \lambda=1.$$
$$P=(-2,-3,5) \implies |a|+|b|+|c|=2+3+5=10.$$

Solutions · 1