Match the following lists:b.  The point of intersection of the line x−2−3=y−1−2=z−32 and the plane      2x+y-z=3 isc.  The foot of the perpendicular from the point (1, 1,2) to the plane      2x - 2y+42+5=0 isd. The intersection point of the lines x−12=y−23=z−34 and      x−45=y−12=z  is

a.   If the required image is (x, y, z), then x−32=y−51=z−71=−2(6+5+7+18)22+12+12=−12  or (x,y,z)≡(−21,−7,−5)b.  Any point on the line x−2−3=y−12=z−32=λ is (−3λ+2,2λ+1,2λ+3)  which lies on plane 2x + y - z = 3.      Therefore      −6λ+4+2λ+1−2λ−3=3−6λ=1λ=−1/6       Therefore, the point is 52,23,83. c.   If (x, y, z) is the required foot of the perpendicular, then x−12=y−1−2=z−24=−(2−2+8+5)22+(−2)2+42       or  (x,y,z)≡−112,2512,−212d.    Any point on the line x−12=y−23=z−34=λ is P(2λ+1,3λ+2,4λ+3) which satisfies the line x−45=y−12=z1        or 2λ+1−45=3λ+2−12=4λ+31or λ=−1         The required point is (- 1, - 1, - 1)