Determinant of a $$3\times3$$ Matrix

Statement I

Put $$a=\cos\alpha,\;b=\cos\beta,\;c=\cos\gamma$$.
The first determinant becomes

$$\Delta_1=\begin{vmatrix}1&a&b\\a&1&c\\b&c&1\end{vmatrix}$$

For any symmetric matrix of this type the standard expansion gives
$$\Delta_1=1+2abc-a^{2}-b^{2}-c^{2}\;.\;-(1)$$

The second determinant is

$$\Delta_2=\begin{vmatrix}0&a&b\\a&0&c\\b&c&0\end{vmatrix}$$

Expanding along the first row:

$$\Delta_2=0\cdot\Bigl|\begin{smallmatrix}0&c\\c&0\end{smallmatrix}\Bigr| -a\Bigl|\begin{smallmatrix}a&c\\b&0\end{smallmatrix}\Bigr| +b\Bigl|\begin{smallmatrix}a&0\\b&c\end{smallmatrix}\Bigr| = -a(-bc)+b(ac)=2abc\;.\;-(2)$$

Given $$\Delta_1=\Delta_2$$, substitute $$(1)$$ and $$(2)$$:

$$1+2abc-a^{2}-b^{2}-c^{2}=2abc \;\;\Longrightarrow\;\;1-a^{2}-b^{2}-c^{2}=0$$

Hence $$\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma=a^{2}+b^{2}+c^{2}=1$$, not $$\frac{3}{2}$$.
So Statement I is false.

Statement II

Let

$$D(x)=\begin{vmatrix}x^{2}+x & x+1 & x-2\\ 2x^{2}+3x-1 & 3x & 3x-3\\ x^{2}+2x+3 & 2x-1 & 2x-1\end{vmatrix}$$

Use the cofactor rule $$D=A_1E_1-B_1E_2+C_1E_3$$ with
$$(A_1,B_1,C_1)=(x^{2}+x,\;x+1,\;x-2).$$

$$E_1=\begin{vmatrix}0&3x-3\\2x-1&0\end{vmatrix}=3(2x-1)=6x-3$$
$$\;\;\Longrightarrow\;A_1E_1=(x^{2}+x)(6x-3)=6x^{3}+3x^{2}-3x$$

$$E_2=\begin{vmatrix}2x^{2}+3x-1&3x-3\\x^{2}+2x+3&2x-1\end{vmatrix} =x^{3}+x^{2}-8x+10$$
$$\;\;\Longrightarrow\;B_1E_2=(x+1)(x^{3}+x^{2}-8x+10) =x^{4}+2x^{3}-7x^{2}+2x+10$$

$$E_3=\begin{vmatrix}2x^{2}+3x-1&3x\\x^{2}+2x+3&2x-1\end{vmatrix} =x^{3}-2x^{2}-14x+1$$
$$\;\;\Longrightarrow\;C_1E_3=(x-2)(x^{3}-2x^{2}-14x+1) =x^{4}-4x^{3}-10x^{2}+29x-2$$

Therefore

$$D(x)=\bigl(6x^{3}+3x^{2}-3x\bigr) -\bigl(x^{4}+2x^{3}-7x^{2}+2x+10\bigr) +\bigl(x^{4}-4x^{3}-10x^{2}+29x-2\bigr)$$

Simplifying term by term:

$$D(x)=24x-12=12(2x-1)$$

Thus $$p=24,\;q=-12$$ and
$$p^{2}=24^{2}=576,\quad 196q^{2}=196\cdot144=28224\neq576.$$

Hence Statement II is also false.

Both statements are false  →  Option B.

For the system of linear equations to have infinitely many solutions, the determinant of the coefficient matrix must be zero ($$\Delta = 0$$) and the system must be consistent ($$\Delta_z = 0$$). The coefficient matrix $$A$$ is

$$A = \begin{bmatrix} 1 & 1 & 2 \\ 2 & 3 & a \\ -1 & -3 & b \end{bmatrix}$$

so its determinant is computed by expanding along the first row:

$$\Delta = 1 \cdot \det \begin{bmatrix} 3 & a \\ -3 & b \end{bmatrix} - 1 \cdot \det \begin{bmatrix} 2 & a \\ -1 & b \end{bmatrix} + 2 \cdot \det \begin{bmatrix} 2 & 3 \\ -1 & -3 \end{bmatrix}$$

Computing each $$2 \times 2$$ determinant gives:

$$\Delta = 1(3b + 3a) - 1(2b + a) + 2(-6 + 3) = 2a + b - 6$$

Setting $$\Delta = 0$$ gives $$2a + b = 6$$. Next, computing $$\Delta_z$$ by replacing the third column with the constant terms:

$$\Delta_z = \det \begin{bmatrix} 1 & 1 & 6 \\ 2 & 3 & a + 1 \\ -1 & -3 & 2b \end{bmatrix}$$

Expanding along the first row yields:

$$\Delta_z = 1 \cdot (6b + 3a + 3) - 1 \cdot (4b + a + 1) + 6 \cdot (-3) = 2a + 2b - 16$$

Setting $$\Delta_z = 0$$ gives $$2a + 2b = 16$$, which simplifies to $$a + b = 8$$.

We now have a system of two equations, $$2a + b = 6$$ and $$a + b = 8$$. Subtracting the second equation from the first gives $$a = -2$$. Substituting this value into the second equation yields $$b = 10$$.

Finally, substituting these values into the required expression:

$$7a + 3b = 7(-2) + 3(10) = -14 + 30 = 16$$

Therefore, the final answer is $$16$$.

The system: $$x+y+z=6$$, $$x+2y+5z=9$$, $$x+5y+\lambda z=\mu$$ has no solution when the determinant of coefficients is zero but the system is inconsistent.

Compute the determinant of the coefficient matrix: $$D = \begin{vmatrix}1&1&1\\1&2&5\\1&5&\lambda\end{vmatrix} = 1(2\lambda-25)-1(\lambda-5)+1(5-2) = 2\lambda-25-\lambda+5+3 = \lambda-17$$.

For no solution one requires $$D = 0 \implies \lambda = 17$$.

With $$\lambda = 17$$, the determinant of the augmented matrix replacing the first column by the constants becomes $$D_x = \begin{vmatrix}6&1&1\\9&2&5\\\mu&5&17\end{vmatrix} = 6(34-25)-1(153-5\mu)+1(45-2\mu) = 54-153+5\mu+45-2\mu = 3\mu-54$$.

Inconsistency requires $$D_x \neq 0 \implies 3\mu \neq 54 \implies \mu \neq 18$$.

Thus the system has no solution when $$\lambda = 17$$ and $$\mu \neq 18$$.

The correct answer is Option 3: $$\lambda = 17, \mu \neq 18$$.