Let the intersection point of l2 and l3 be C(x,y,z)

Since they are perpendicular, they must intersect at right angles. Their direction vectors must be perpendicular as well. The dot product of these two vectors equals 0.

Find a direction vector for l3 first.
Since B(3,-1,0) is on l3 as well, we can do the following:

A direction vector parallel to l3 is : Vector BC

= Vector OC - Vector OB

(x - 3, y - (-1), z - 0)
= (x - 3, y + 1, z)

The dot product of these two vectors equals 0, so:

(x-3,y+1,z)·(0.5,1,2) = 0

0.5x - 1.5 + y + 1 + 2z = 0
0.5x + y + 2z = 0.5
x + 2y + 4z = 1

We get an equation of a plane. This plane contains l3 and points B and C.

Now we just have to solve for the intersection of this plane and l2.

Recall the Cartesian equation of l2 that was given :

2x = y + 7 = (z-3)/2

From this,
2x = y + 7
y = 2x - 7

(z - 3)/2 = 2x
z - 3 = 4x
z = 4x + 3

Sub y = 2x - 7 and z = 4x + 3 into the equation of the plane,

x + 2(2x - 7) + 4(4x + 3) = 1
x + 4x - 14 + 16x + 12 = 1
21x = 3
x = 3/21
x = 1/7

So,
y = 2(1/7) - 7 = -47/7
z = 4(1/7) + 3 = 25/7

(1/7,-47/7,25/7) is the intersection point of l2 and l3.

Thanks !!! This question was asked before eqn of plane was taught

Welcome. But anyway, we don't actually need to know about the concept of an equation of a plane in order to solve it since it's mainly solving simultaneous equations in 3 variables haha.

The plane part is more about the conceptual understanding or geometric interpretation.

The Cartesian equation of l2 is convenient here as it allows for substitution of the variables such that the three-variable equation can be reduced to a single-variable one.

As l2 and l3 intersect at C, the x,y, and z values of C must satisfy both equations.

Good job, as usual, J.