Show that the lines and 3x – 2y + z + 5 = 0 = 2x + 3y + 4z – 4 intersect. Find the equation of the plane in which they lie and also their point of intersection.

we have, equation of the line is

General point on the line is given by (3 – 4, 5 – 6, – 2 + 1) …… (1)

Another equation of line is

3x – 2y + z + 5 = 0

2x + 3y + 4z – 4 = 0

Let a, b, c be the direction ratio of the line so, it will be perpendicular to normal of 3x – 2y + z + 5 = 0 and 2x + 3y + 4z – 4 = 0

So, using a1a2 + b1b2 + c1c2 = 0

(3)(a) + (– 2)(b) + (1)(c) = 0

3a – 2b + c = 0 …… (2)

Again, (2)(a) + (3)(b) + (4)(c) = 0

2a + 3b + 4c = 0 …… (3)

Solving (2) and (3) by cross – multiplication,

Direction ratios are proportional to – 11, – 10, 13

Let z = 0 so

3x – 2y = – 5 …… (i)

2x + 3y = 4 …… (ii)

Solving (i) and (ii) by eliminations method,

– 13y = – 22

Put y in equation (i)

3x – 2y = – 5

3x – 2 = – 5

3x – = – 5

3x = – 5 +

3x =

x =

so, the equation of the line (2) in symmetrical form,

Put the general point of a line from equation (1)

The equation of the plane is 45x – 17y + 25z + 53 = 0

Their point of intersection is (2, 4, – 3)