X is `5/8` of y .y is `3/4` of z and z is `2/5` of k .what part of k is x?

The planes 2x - y + 4z = 3 and 5x - 2.5y + 10z = 6 are :

If the line (x-2)/-1=(y+2)/1=(z+k)/4 is one of the angle bisector of the lines x/1=y/-2=z/3 and x/-2=y/3=z/1 then the value of k is

If x : y = 1 : 3, y :z = 5 :k, z : t = 2:5 and t:x = 3:4 , then what is the value of k ?

Subtract the sum of 5x - 4y + 6z and -8x + y - 2z from the sum of 12x - y + 3z and -3x + 5y - 8z

if the lines (x - 1)/(-3) = ( y - 2)/(2k) = ( z -3)/(2) and (x -1) /(3k) = ( y - 5)/(1) = (z - 6 ) /(-5) are at night angle , then find the value of k .

Find the angle between the following pairs of lines : (i) vec(r) = 3 hat(i) + 2 hat(j) - 4 hat(k) + lambda (hat(i) + 2 hat(j) + 2 hat(k) ) . vec(r) = 5 hat(j) - 2 hat(k) + mu ( 3 hat(i) + 2 hat(j) + 6 hat(k)) (ii) vec(r) = 3 hat(i) + hat(j) - 2 hat(k) + lambda (hat(i) - hat(j) - 2 hat(k) ) . vec(r) =(2 hat(i) - hat(j) - 56 hat(k)) + mu ( 3 hat(i) - 5 hat(j) - 4 hat(k)) (iii) (x-2)/(2) = (y - 1)/(5) = (z + 3)/(-3) and (x + 2)/(-1) = (y - 4)/(8) = (z - 5)/(4) (iv) (x - 4)/(3) = (y + 1)/(4) = (z - 6)/(5) and (x-5)/(1) = (2y +5)/(-2) = (z - 3)/(1) (v) (5 -x)/(3) = (y + 3)/(-4) , z = 7 and x = (1-y)/(2) = (z - 6)/(2) (vi) (x + 3)/(3) = (y - 1)/(5) = (z + 3)/(4) and (x + 1)/(1) = (y - 4)/(1) = (z-5)/(2) .

Find the angle between the planes : (i) 3 x - 6y - 2z = 7 " " 2x + y - 2z = 5 (ii) 4 x + 8y + z = 8 " " and y + z = 4 (iii) 2 x - y + z =6 " " and x + y + 2z = 7 .