JEE Mathematics 2023 Question with Solution

Given: The printed equation contains $yz$, but the solution interprets this as the coefficient of $z$. So we compare the required plane with

$$\alpha x + \beta y + \gamma z = 1$$

passing through $(3, -2, 5)$ and perpendicular to the line joining $(1, 2, 3)$ and $(-2, 3, 5)$.

Find: $\alpha \beta \gamma$.

The direction vector of the given line is

$$\vec d = (-2-1)\mathbf{i} + (3-2)\mathbf{j} + (5-3)\mathbf{k} = -3\mathbf{i} + \mathbf{j} + 2\mathbf{k}$$

A plane perpendicular to this line has a normal vector parallel to this direction vector. Taking an equivalent normal vector,

$$\vec n = 3\mathbf{i} - \mathbf{j} - 2\mathbf{k}$$

Hence the plane can be written as

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

Since it passes through $(3, -2, 5)$,

$$3(3) - (-2) - 2(5) + \lambda = 0$$ $$9 + 2 - 10 + \lambda = 0$$ $$\lambda = -1$$

Therefore, the plane is

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

Comparing with

$$\alpha x + \beta y + \gamma z = 1$$

we get

$$\alpha = 3, \quad \beta = -1, \quad \gamma = -2$$

So,

$$\alpha \beta \gamma = 3(-1)(-2) = 6$$

Therefore, the required value is $6$.

The expression $yz$ makes the given equation non-linear, so it cannot represent a plane. The extracted solution treats this as a typographical issue and uses

$$\alpha x + \beta y + \gamma z = 1$$

instead. Under that interpretation, the coefficients are obtained from the plane equation and the final product becomes $6$.