How to find the point where the line \[x = - 1 - t,y = 2 + t,z = 1 + t\] intersects the plane \[3x + y + 3z = 1\] ?

Hint:
Here we will find the value of the coordinate of the points where the given three lines intersect the given plane. As the given line intersects the plane so the value of the coordinate will satisfy the equation of the plane, so we will substitute the value of the points from the line into the plane and get the value of the variable \[t\]. Then we will substitute the value of the variable back in the equation of line to get the value of the coordinate to get the required answer.

Complete step by step solution:
The three lines given to us are
\[x = - 1 - t\]……\[\left( 1 \right)\]
\[y = 2 + t\]……\[\left( 2 \right)\]
\[z = 1 + t\]…….\[\left( 3 \right)\]
The plane at which the above line intersect is,
\[3x + y + 3z = 1\]
Substituting the value from equation \[\left( 1 \right)\], \[\left( 2 \right)\] and \[\left( 3 \right)\] in above value we get,
\[ \Rightarrow 3\left( { - 1 - t} \right) + \left( {2 + t} \right) + 3\left( {1 + t} \right) = 1\]
\[\begin{array}{l} \Rightarrow - 3 - 3t + 2 + t + 3 + 3t = 1\\ \Rightarrow 2 + t = 1\end{array}\]
Keeping \[t\] term on one side and taking rest to the other we get,
\[\begin{array}{l} \Rightarrow t = 1 - 2\\ \Rightarrow t = - 1\end{array}\]
So, we get the value of \[t\] as -1.
Substituting value of \[t\] in equation \[\left( 1 \right)\], \[\left( 2 \right)\] and \[\left( 3 \right)\], we get
\[x = - 1 - \left( { - 1} \right) = - 1 + 1 = 0\]
$y=2-1=1$
\[z = 1 - 1 = 0\]
So, we get our point as,

\[\left( {x,y,z} \right) = \left( {0,1,0} \right)\]

Note:
A plane is a two-dimensional surface that can extend to infinity on either direction. It can be anything, a line, a circle of even a triangle. Geometry of a plane is all about the shapes that are on a flat surface. When line and plane intersect the result can be either a line or a point or it can be an empty set. When a line intersects a plane all points of the line lie in the plane as well.