Can you type up the work on these pictures so I can paste it onto a doc.

1)

\fwe have . to determine the equation of the line that perpendicular to the plane gz-14+7:1. passing through (2, 3. -D). . we know that, aline passing hough (abe) and the direction malian (k.Im). Mop Since the Line is prev perdicalento 3x-274731 Hence the direction ratio is the normal of the plant Now normal of the plane is (3,-2, 1). Ha vee (o.bix) = (23-3) (K LM ) = (3,-2 1) Equation of the line 2+3 = 3 - 3 -2 1 Y - 3 2+3 21 1 3 2 = 2+36, y= 3-26, = -3+. pana metric form is 567 2 43t 2 :-3 # vector form is "= (23t, 3+ 21, 3 # t ) = (213,-3) ++(8,2,1)The given equation is: - ty - 2 = 1 2x -54 + 2 = -1 Th- 7y - 2 = K. We can write above equation in augmented matsin formas i- 5 K Now, using clementry soo operation - L R, - R - 2RX -1 K R3 - Rg - TR. - 7 - 3 RX - Rg - 2Rg - 73 - 3 0 - 14 6 K- 7 0-73 3 0 0 0 K-L . . If k = 1 last sow ( Ry) is all a then , we will have tofinally infinite solution So. when K=1 last mal (Rg) equal to to ool of Hence it is a line.. So Verton equation off line is. "= any Point on the lineThe distance of the point POLly, 2, ) from the plane au+ by + ez +d = 0 1's P Equation of the plane is 3x - 44 - 127 - 8 = 0 and the point ""SAC - 1 1/ 2 ) ." The distance fromAll,?) to the plane qx - 44 - 12 7- 8 = 0 135 3 ( - 17 - 4017 - 12 (2) -8 V 37+4+123 - 39 3 13 the distance from B( 3,1-2) to the plane 8k-84+42-7=0 8 x3 -8* 1 + 4(-2)-7 8 +8 +42 = 1 14 4 12