 Hello friends, let's discuss the following problem. It says, find the distance of the point minus 1, minus 5, minus 10. From the point of intersection of the line, vector r is equal to 2i cap minus j cap plus 2k cap plus lambda times the vector 3i cap plus 4j cap plus 2k cap and the plane vector r dot i cap minus j cap plus k cap is equal to 5. Let us now proceed on with the solution. Equation of line is given to be vector r is equal to 2i cap minus j cap plus 2k cap plus lambda times the vector 3i cap plus 4j cap plus 2k cap and in the Cartesian form equation of line is given by x minus 2 upon 3 is equal to y minus minus 1 upon 4 is equal to z minus 2 upon 2. Let's call this as lambda. So, from this we have x is equal to 3 lambda plus 2, y is equal to 4 lambda minus 1 and z is equal to 2 lambda minus its plus 2. So, the coordinates of any point on the line 1 are given by 3 lambda plus 2 comma 4 lambda minus 1 comma 2 lambda plus 2. Now, we are given that the line and the plane intersect. So, if line and plane intersect, this point must lie on the plane. Now, the equation of plane is given to be vector r dot i cap minus j cap plus k cap is equal to 5 and vector r is the position vector of any arbitrary point on the plane. So, it is x i cap plus y j cap plus z k cap dot i cap minus j cap plus k cap is equal to 5. Now, we can write the Cartesian form of equation taking the dot product of these two vectors. We get the Cartesian form it is given by x minus y plus z is equal to 5 which is again equal to x minus y plus z minus 5 is equal to 0. Let's call this as 2 and let's call the coordinate of any point on the line as a. Now, this point lies on the plane that means a satisfies 2. Therefore, 3 lambda plus 2 minus 4 lambda minus 1 plus 2 lambda plus 2 minus 5 must be equal to 0. This implies 3 lambda plus 2 minus 4 lambda plus 1 plus 2 lambda plus 2 minus 5 is equal to 0. This implies 5 lambda minus 4 lambda plus 5 minus 5 is equal to 0 and this implies lambda is equal to 0. Now, put lambda is equal to 0 in a and we get the coordinates as 3 into 0 plus 2 4 into 0 minus 1 2 into 0 plus 2 and the coordinates are 2 minus 1 2 and this is the point of intersection of line and the plane. Now, we have to find the distance between the point minus 1 minus 5 minus 10 and the point of intersection of the line and the plane. Now, the distance between the point minus 1 minus 5 minus 10 and the point 2 minus 1 2 is given by root of minus minus 1 whole square plus minus 1 minus of minus 5 whole square plus 2 minus minus 10 whole square and it is equal to root of 9 plus 16 plus 144 which is equal to root of 169 which is equal to 13. So, the distance between these two points is 13 units. Hence, the answer is 13 units. So, this completes the question. Why for now take care. Have a good day.