a)  You are given the line "parametrized by the parameter t".


    In order to verify that the point (2,-1,3) lies on L,  you need to show that there is such "t" that


         2 = -t + 2,     (1)
        -1 = 2t - 1,     (2)
         3 =  t + 3.     (3)


    To do it,  from equation (1)  find  t = -2 + 2 = 0  and check that this value of t satisfies equations (2) and (3).

    You can easily check it on your own.


    So, this question is answered.



    Next question relates to the point (1,1,1),  and we need to show that THERE IS NO such t which satisfies the equations

         1 = -t + 2,     (4)
         1 = 2t - 1,     (5)
         1 =  t + 3.     (6)


    Indeed, from equation (4) t = -1 + 2 = 1.  The value of t= 1 still satisfies equation (5), but does not satisfy equation (6).


     So, this question is answered, too.



b)  The straight line L in   has the guiding vector (-1,2,1) comprised of the coefficients of its parametrization.


    The orthogonal plane to this vector has the form 


        -1x + 2y + 1z = c,


     where "c" is some constant, whose value we determine using the condition that the point (2,-1,3) belongs to the plane.


     So, to find "c", we simply substitute the coordinates  x= 2, y= -1, z= 3 into equation (*) to get

        c = (-1)*2 + 2*(-1) + 1*3 = -2 -2 + 3 = -1.


    Thus your plane under the question b) has the equation

        -x + 2y + z = -1.