I have 2 Question on $3-D$ Geometry
(1) The point on the Line $\displaystyle \frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$ which is Nearest to the Line
$\displaystyle \frac{x+1}{7}=\frac{y+1}{6}=\frac{z+1}{1}$ is
(2) If a Plane Contain $3-$ Lines drawn through $(1,1,1)$ and has a direction Ratios
$(1,-4,-1)\;\;,(3,5,7)$ and $(2,9,\mu)$. Then value of $\mu=$
For part 1, see this page.
For the second part: The equation of the three lines are $$\frac{x-1}{l_i}+\frac{y-1}{m_i}+ \frac{z-1}{n_i}, i = 1,2,3 $$ where $(l_i,m_i,n_i)$ is the direction ratio for the line.
Let the equation of the plane containing these three lines be $ax+by+cz =1 $. The plane contains the point $(1,1,1)$. So, $$a+b+c =1 $$ Also, $(a,b,c)$ is the direction ratio of the normal to the plane. For any line in the plane with direction ratio $(l,m,n)$$$al+bm+cn = 0$$
So, $$a.1+b.(-4)+c.(-1) =0$$ $$a.3+b.5+c.7 =0$$ Use these three equations to find out $a,b,c$. Then, use $a.2+b.9+c.\mu =0$ to find out $\mu$.