Question

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Answer 1

Given the equation of line

$\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12} = p(let)$

where p is a point of intersection

x = 3p + 2

y = 4p -1

z = 12p +2

so point of intersection D(3p+2 ,4p - 1,12p+2)

Now it is given that point of intersections of line to plane is

x - y + z = 16:

the point D satisfy the line so on putting the value of x,y and z we get

3p + 2 - 4p + 1 + 12p + 2 = 16

11p = 11

p = 1

Now putting the value of p in point D we get

D (5,3,14)

Now distance between the point (1,0,2) and (5,3,4) is

S =

$\sqrt{ {(5 - 1)}^{2} + {(3 - 0)}^{2} + {(14 - 2)}^{2} }$

S =

$\sqrt{16 + 9 + 144}$

S =

$\sqrt{169 }$

13

Answer 2

hope it will help you...........