Solution:

(a) It is given that,
The points are $(1,1,-1),(6,4,-5),(-4,-2,3)$.
Let,
$\begin{array}{l}
=\left|\begin{array}{ccc}
1 & 1 & -1 \\
6 & 4 & -5 \\
-4 & -2 & 3
\end{array}\right| \\
=1(12-10)-1(18-20)-1(-12+16) \\
=2+2-4 \\
=0
\end{array}$
Since, the value of determinant is 0 .
As a result, the points are collinear as there will be infinite planes passing through the given 3 points.

(b) $(1,1,0),(1,2,1),(-2,2,-1)$
The given points are $(1,1,0),(1,2,1),(-2,2,-1)$.
Let,
$\begin{array}{l}
=\left|\begin{array}{ccc}
1 & 1 & 0 \\
1 & 2 & 1 \\
-2 & 2 & -1
\end{array}\right| \\
=1(-2-2)-1(-1+2) \\
=-4-1 \\
=-5 \neq 0
\end{array}$
As, there passes a unique plane from the given 3 points.
Eq. of the plane passing through the points,
$\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right),\left(\mathrm{x}_{2}, \mathrm{y}_{2}, \mathrm{z}_{2}\right)$ and $\left(\mathrm{x}_{3}\right.$, $\left.\mathrm{y}_{3}, \mathrm{z}_{3}\right)$, i.e.,
$=\left|\begin{array}{lll}
x-x_{1} & y-y_{1} & z-z_{1} \\
x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\
x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1}
\end{array}\right|$
Substitute the values and simplify
$\begin{array}{l}
=\left|\begin{array}{ccc}
\mathrm{x}-1 & \mathrm{y}-1 & \mathrm{z} \\
\mathrm{x}_{2}-1 & \mathrm{y}_{2}-1 & \mathrm{z}_{2} \\
\mathrm{x}_{3}-1 & \mathrm{y}_{3}-1 & \mathrm{z}_{3}
\end{array}\right| \\
=\left|\begin{array}{ccc}
\mathrm{x}-1 & \mathrm{y}-1 & \mathrm{z} \\
1-1 & 2-1 & 1 \\
-2-1 & 2-1 & -1
\end{array}\right| \\
=\left|\begin{array}{ccc}
\mathrm{x}-1 & \mathrm{y}-1 & \mathrm{z} \\
0 & 1 & 1 \\
-3 & 1 & -1
\end{array}\right| \\
=(\mathrm{x}-1)(-2)-(\mathrm{y}-1)(3)+3 \mathrm{z}=0 \\
\Rightarrow-2 \mathrm{x}+2-3 \mathrm{y}+3+3 \mathrm{z}=0 \\
=2 \mathrm{x}+3 \mathrm{y}-3 \mathrm{z}=5
\end{array}$
As a result, the required equation of the plane is $2 x+3 y-3 z=5$.