Answer
(a) $x+y+z=c$
This is the family of planes that have normal vectors parallel to $ \lt 1,1,1\gt$. The distance from the plane to the origin varies by changing the value of $c$. For positive c values, the planes intersect to form an equilateral triangle in the first octant. For negative c values, the equilateral triangle is formed with the octant opposite the first.
(b) $x+y+cz=1$
This is the family of planes that intercects the $xy$-plane at the 2D line $x+y=1$. With $c=0$, the plane is parallel to the z-axis. For larger $c$ values, the plane moves closer to the xy-plane.
(c) $y cos \theta+zsin \theta=1$
Because there is no $x$, the planes will be parallel to the $x-$ axis.
These lines are basically tangent lines to a circle of radius $1$.
So, the planes are tangent lines to a circle of radius $1$ on the $yz$ plane extended in the direction parallel to the $x-axis$.
Work Step by Step
(a) $x+y+z=c$
This is the family of planes that have normal vectors parallel to $ \lt 1,1,1\gt$. The distance from the plane to the origin varies by changing the value of $c$. For positive c values, the planes intersect to form an equilateral triangle in the first octant. For negative c values, the equilateral triangle is formed with the octant opposite the first.
(b) $x+y+cz=1$
This is the family of planes that intercects the $xy$-plane at the 2D line $x+y=1$. With $c=0$, the plane is parallel to the z-axis. For larger $c$ values, the plane moves closer to the xy-plane.
(c) $y cos \theta+zsin \theta=1$
Because there is no $x$, the planes will be parallel to the $x-$ axis.
These lines are basically tangent lines to a circle of radius $1$.
So, the planes are tangent lines to a circle of radius $1$ on the $yz$ plane extended in the direction parallel to the $x-axis$.
