Find the shortest distance between the z-axis and the line, `x+y+2z-3=0,2x+3y+4z-4=0.`

Explanation

let `p_1` and `p_2` be the points on the given planes having coordinates (1,1,2) & (2,3,4) respectively

R = `|(i,j,k),(1,1,2),(2,3,4)|`

`=i(-2)-j(0) + k(1) = -2i+k`

`(-2,0,1)`

if z=0 `x+y-3 = 0 & 2x+3y-4 = 0`

we get `2(3-y) + 3y-4 = 0`

then, `y=-2 & x = 5`

so,intersection point `(5,-2,0)`

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