The system of linear equations
$$\lambda x + 2y + 2z = 5$$
$$2\lambda x + 3y + 5z = 8$$
$$4x + \lambda y + 6z = 10$$ has

The three equations can be written in matrix form as $$A\mathbf x=\mathbf b$$ where

$$A=\begin{bmatrix}\lambda&2&2\\[2pt]2\lambda&3&5\\[2pt]4&\lambda&6\end{bmatrix},$$ $$\mathbf x=\begin{bmatrix}x\\y\\z\end{bmatrix},$$ $$\mathbf b=\begin{bmatrix}5\\8\\10\end{bmatrix}.$$

For a system of three linear equations, the basic facts are:

• If the determinant $$\det A\ne0,$$ the system has a unique solution.
• If $$\det A=0,$$ we must compare the rank of $$A$$ with the rank of the augmented matrix $$[A\;|\;\mathbf b]$$. Equal ranks give infinitely many solutions, while unequal ranks give no solution.

We therefore begin by finding $$\det A$$. Using the first row expansion formula

$$\det A =\lambda\begin{vmatrix}3&5\\ \lambda&6\end{vmatrix} -2\begin{vmatrix}2\lambda&5\\ 4&6\end{vmatrix} +2\begin{vmatrix}2\lambda&3\\ 4&\lambda\end{vmatrix}.$$

Evaluating each $$2\times2$$ determinant one by one, we have

$$\begin{aligned} \det A &= \lambda(3\cdot6-5\lambda) -2(2\lambda\cdot6-5\cdot4) +2(2\lambda\cdot\lambda-3\cdot4)\\[4pt] &=\lambda(18-5\lambda)-2(12\lambda-20)+2(2\lambda^{2}-12). \end{aligned}$$

Multiplying out and collecting like terms gives

$$\begin{aligned} \det A &=(18\lambda-5\lambda^{2})-24\lambda+40+4\lambda^{2}-24\\[4pt] &=-\lambda^{2}-6\lambda+16\\[4pt] &=-(\lambda^{2}+6\lambda-16). \end{aligned}$$

Setting $$\det A=0$$ yields

$$\lambda^{2}+6\lambda-16=0\quad\Longrightarrow\quad (\lambda-2)(\lambda+8)=0,$$

so $$\lambda=2\text{ or }\lambda=-8$$ are the only values for which the determinant vanishes. For every $$\lambda$$ other than these two, $$\det A\neq0$$ and the system has exactly one solution.

We now analyse the critical values.

Case 1: $$\lambda=2$$.

Substituting $$\lambda=2$$ in the equations gives

$$\begin{aligned} 2x+2y+2z&=5\qquad&(1)\\ 4x+3y+5z&=8\qquad&(2)\\ 4x+2y+6z&=10\qquad&(3) \end{aligned}$$

Multiply equation (1) by $$2$$:

$$4x+4y+4z=10\qquad(1')$$

Subtracting (1′) from (3) eliminates $$x$$:

$$[4x+2y+6z]-[4x+4y+4z]=10-10\;\;\Longrightarrow\;\;-2y+2z=0,$$

so $$-y+z=0\;\Longrightarrow\;z=y.$$

Subtracting equation (2) from (1′) also eliminates $$x$$:

$$[4x+4y+4z]-[4x+3y+5z]=10-8\;\;\Longrightarrow\;\;y-z=2.$$

But with $$z=y$$, the relation $$y-z=2$$ becomes $$0=2,$$ a contradiction. Therefore the augmented matrix has rank 3 while the coefficient matrix has rank 2, so the system is inconsistent. There is no solution when $$\lambda=2$$.

Case 2: $$\lambda=-8$$.

Substituting $$\lambda=-8$$ in the equations yields

$$\begin{aligned} -8x+2y+2z&=5\qquad&(4)\\ -16x+3y+5z&=8\qquad&(5)\\ 4x-8y+6z&=10\qquad&(6) \end{aligned}$$

Double equation (4): $$-16x+4y+4z=10\;(4′).$$ Subtract (5) from (4′):

$$(-16x+4y+4z)-(-16x+3y+5z)=10-8\;\;\Longrightarrow\;\;y-z=2.$$

Again from (4): $$-8x+2y+2z=5\;\Longrightarrow\;x=\dfrac{-5+2y+2z}{8}.$$ Substituting $$z=y-2$$ (from $$y-z=2$$) in this expression gives $$x=\dfrac{-9+4y}{8}.$$ Replacing $$x$$ and $$z$$ in equation (6) finally leads to the contradiction $$-\dfrac{33}{2}=10.$$ Thus the system is also inconsistent when $$\lambda=-8.$$ There is no solution here as well, certainly not a unique one.

Summarising:

• For $$\lambda=2$$: determinant zero and the system is inconsistent → no solution.
• For $$\lambda=-8$$: determinant zero and the system is inconsistent → no solution.
• For every other $$\lambda$$: determinant non-zero → exactly one solution.

Among the given options, only Option C correctly states a situation that actually occurs.

Hence, the correct answer is Option C.