JEE Mathematics 2023 Question with Solution

Given: The required plane passes through the line of intersection of

$$x - 2y - z - 5 = 0$$

and

$$x + y + 3z - 5 = 0$$

It is parallel to the line of intersection of

$$x + y + 2z - 7 = 0$$

and

$$2x + 3y + z - 2 = 0$$

Also, the plane is $ax + by + cz = 65$.

Find: The distance of the point $(a,b,c)$ from the plane

$$2x + 2y - z + 16 = 0$$

A plane through the intersection of the first two planes is

$$(x - 2y - z - 5) + b(x + y + 3z - 5) = 0$$

So its equation becomes

$$(1+b)x + (-2+b)y + (-1+3b)z - 5(1+b) = 0$$

Since this is the same plane as $ax + by + cz = 65$, comparing with the solution working gives

$$a = 1+b, \quad b = -2+b, \quad c = -1+3b$$

Using the parallelism condition with the second line, the determinant is zero:

$$\begin{vmatrix} 1+b & -2+b & -1+3b \\ 1 & 1 & 2 \\ 2 & 3 & 1 \end{vmatrix} = 0$$

This gives

$$b = 12$$

Therefore the plane is

$$13x + 10y + 35z = 65$$

Hence

$$(a,b,c) = (13,10,35)$$

Now distance of the point $(13,10,35)$ from the plane

$$2x + 2y - z + 16 = 0$$

is

$$d = \frac{|2(13) + 2(10) - 35 + 16|}{\sqrt{2^2 + 2^2 + (-1)^2}}$$ $$d = \frac{|26 + 20 - 35 + 16|}{\sqrt{9}} = \frac{27}{3} = 9$$

Therefore, the distance is $9$.

Given: The required plane contains the first line and is parallel to the second line.

Find: The distance asked in the question.

The first line is the intersection of planes with normals

$$\vec{n}_1 = (1,-2,-1), \quad \vec{n}_2 = (1,1,3)$$

So a family of planes through that line is obtained by

$$(x - 2y - z - 5) + \lambda (x + y + 3z - 5) = 0$$

Its normal vector is

$$(1+\lambda, -2+\lambda, -1+3\lambda)$$

The second line is the intersection of planes

$$x + y + 2z - 7 = 0, \quad 2x + 3y + z - 2 = 0$$

so its direction vector is perpendicular to both normals $(1,1,2)$ and $(2,3,1)$. For the required plane to be parallel to this line, the plane normal must be perpendicular to that direction, which is exactly the determinant condition used in the solution:

$$\begin{vmatrix} 1+\lambda & -2+\lambda & -1+3\lambda \\ 1 & 1 & 2 \\ 2 & 3 & 1 \end{vmatrix} = 0$$

From the extracted solution,

$$\lambda = 12$$

Thus the plane is

$$13x + 10y + 35z = 65$$

So

$$(a,b,c) = (13,10,35)$$

Using point-to-plane distance formula from $(13,10,35)$ to

$$2x + 2y - z + 16 = 0$$

we get

$$\frac{|2\cdot 13 + 2\cdot 10 - 35 + 16|}{\sqrt{4+4+1}} = \frac{27}{3} = 9$$

Therefore, the required distance is $9$.