JEE Mathematics 2023 Question with Solution

Given: The required plane passes through $(1,1,2)$. The given line is the intersection of the planes

$$x - 3y + 2z - 1 = 0$$

and

$$4x - y + z = 0$$

Find: The value of $140(C-B+A)$ when the plane is written as $Ax + By + Cz = 1$.

The normals to the two given planes are

$$\vec{n}_1 = \langle 1,-3,2 \rangle, \quad \vec{n}_2 = \langle 4,-1,1 \rangle$$

The direction ratios of the given line are given by the cross product:

$$\vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 4 & -1 & 1 \end{vmatrix} = -\hat{i} + 7\hat{j} + 11\hat{k}$$

So the required plane, being perpendicular to this line, has normal vector with direction ratios

$$\langle -1, 7, 11 \rangle$$

Using the point-normal form through $(1,1,2)$,

$$-1(x-1) + 7(y-1) + 11(z-2) = 0$$

Simplifying,

$$-x + 7y + 11z = 28$$

Now divide throughout by $28$ to write it in the form $Ax + By + Cz = 1$:

$$\frac{-1}{28}x + \frac{7}{28}y + \frac{11}{28}z = 1$$

Hence,

$$A = \frac{-1}{28}, \quad B = \frac{7}{28}, \quad C = \frac{11}{28}$$

Now compute

$$140(C-B+A) = 140\left(\frac{11}{28} - \frac{7}{28} - \frac{1}{28}\right) = 140\left(\frac{3}{28}\right) = 15$$

Therefore, the required value is $15$.

Given: The plane is perpendicular to the given line and passes through $(1,1,2)$. Find: $140(C-B+A)$.

A plane perpendicular to a line has its normal parallel to the line. So first find the direction ratios of the line formed by intersection of

$$x - 3y + 2z - 1 = 0$$

and

$$4x - y + z = 0$$

That direction vector is

$$\langle 1,-3,2 \rangle \times \langle 4,-1,1 \rangle = \langle -1,7,11 \rangle$$

So the plane is

$$-x + 7y + 11z = d$$

Since it passes through $(1,1,2)$,

$$d = -1 + 7 + 22 = 28$$

Thus

$$-x + 7y + 11z = 28$$

Comparing after dividing by $28$,

$$A = -\frac{1}{28}, \quad B = \frac{1}{4}, \quad C = \frac{11}{28}$$

Therefore,

$$140(C-B+A) = 140\left(\frac{11}{28} - \frac{7}{28} - \frac{1}{28}\right) = 15$$

Therefore, the required value is $15$.