If a plane passes through the points (−1,k,0),(2,k,−1),(1,1,2) and is parallel to the line 1x−1=22y+1=−1z+1, then the value of (k−1)(k−2)k2+1 is:
Solutions · 1
Official solution
Rewrite the line as 1x−1=1y+21=−1z+1, direction (1,1,−1).
C(1,1,2); CA=(−2,k−1,−2), CB=(1,k−1,−3).
\[\vec{CA}\times\vec{CB}=(1-k,-8,3-3k).\]
Parallel to the given line means this normal is perpendicular to (1,1,−1):
\[(1-k)-8-(3-3k)=0 \implies 2k=10 \implies k=5.\]
\[\frac{k^2+1}{(k-1)(k-2)}=\frac{26}{4\times3}=\frac{13}{6}.\]