Q26 / 57 · JEE Advanced 2021 (Paper 1) · Mathematics

Let $\alpha, \beta$ and $\gamma$ be real numbers such that the system of linear equations
$$<br>\begin{array}{c}<br>x+2 y+3 z=\alpha \\<br>4 x+5 y+6 z=\beta \\<br>7 x+8 y+9 z=\gamma-1<br>\end{array}<br>$$
is consistent. Let $|M|$ represent the determinant of the matrix
$$<br>M=\left[\begin{array}{ccc}<br>\alpha & 2 & \gamma \\<br>\beta & 1 & 0 \\<br>-1 & 0 & 1<br>\end{array}\right]<br>$$
Let $P$ be the plane containing all those $(\alpha, \beta, \gamma)$ for which the above system of linear equations is consistent, and $D$ be the square of the distance of the point $(0,1,0)$ from the plane $P$.

The value of |M| is

Correct answer: 1

Solution

x+2y+3z=α

4x+5y+6z=β

7x+8y+9z=γ-1

The system of linear equations is consistent it means it has unique solution or infinite solutions

Here, Δ=123456789=0

Hence, there are infinitely many solutions.

If equations have infinitely many solutions, then the equations are linearly connected, i.e., L1+λL2=L3

⇒x+2y+3z-α+λ(4x+5y+6z-β)=7x+8y+9z-γ+1

1+4λ7=2+5λ8=3+6λ9=α+λβγ-1

1+4λ7=2+5λ8⇒λ=-2

Also,1+4λ7=α+λBγ-1

⇒  -1=α-2βr-1

⇒  α-2β+γ=1

Now, M=α2γβ10-101

=α+2(-β)+γ(1)=α-2β+γ=1

Hence, the correct answer is 1.

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