Find t such that vec a=\ln(t-2),-2,6\ , vec b=\t,-2,5\ , vec…

Find $t$ such that $\vec a=\{\ln(t-2),-2,6\}$, $\vec b=\{t,-2,5\}$, $\vec c=\{0,-1,3\}$ are coplanar.

Step 1: Triple product

\[ \begin{vmatrix}\ln(t-2)&-2&6\\t&-2&5\\0&-1&3\end{vmatrix}=\ln(t-2)(-6+5)+2(3t)+6(-t)=-\ln(t-2). \]

The terms with $t$ cancel.

Step 2: Equation and domain

\[ \ln(t-2)=0\ \Longrightarrow\ t-2=1,\qquad t\gt2\ \checkmark \]

\[t=3\]

Check: for $t=3$, $\vec a=\{0,-2,6\}=2\vec c$. The vectors $\vec a$ and $\vec c$ are then collinear, so the three vectors are certainly coplanar (determinant $0+18-18=0$).